Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 1

Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion

Learning Objectives

Upon successful completion of Chapter 8, you will be able to:

• Understand terms.

• State the null and alternative hypotheses.

• Use the 5 steps to test hypotheses for both the critical value and p-value.

• Test specific hypothesized values for means, variances, and proportions.

• Describe the relationship between type I and type II errors.

• Given a value for the test statistic, find the p-value.

I. General Concept

Hypothesis testing or a decision-making process between 2 choices about a population parameter called: the null hypothesis and the alternate or research hypothesis.

A. Kinds of Hypotheses

• The null hypothesis (H0) states that the parameter equals a specific value (or in Chapter 9 two parameters are equal).

• The alternative hypothesis (H1) states the parameter is different from a specific value (or in Chapter 9 the alternate hypothesis is a difference between two parameters).

B. Hypotheses Testing Methods

The traditional or crticial value method which is a location method.

The P-value (prob-value) method which is a comparison or areas method.

C. Common Phrases used in Testing

>

Is greater than, Is increased

<

Is less than, Is decreased or reduced from

≥

Is greater than or equal to, Is at least

≤

Is less than or equal to, Is at most

=

Is equal to, Has not changed

≠

Is not equal to, Has changed from

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D.

H0: Always has the = sign; Northing is different from the usual.

Very Important Notation

H1: Called the research or the alternative hypothesis; includes the direction the researcher hopes to justify, it will always be <, >, or ≠.

E. Stating the Null and Alternate Hypotheses

I.

Calcium is the most abundant mineral in the body and also one of the most important.

It works with phosphorous to build and maintain bones and teeth. The recommended daily allowance (RDA) of calcium is 800 milligrams. Assume we want to test whether people with income below the poverty level receive an average less than the RDA of 800mg.

Example:

.

H0 : μ = 800 H1 : μ < 800 II.

The R.R. Bowker Company of New York collects information on the retail prices of

books. In 2000, the standard deviation of the price of history books was $3.25. Suppose this company wishes to determine whether this year’s standard deviation is higher than the standard deviation price in 2000.

Example:

.

H0 : σ = 3.25 H1 : σ > 3.25

Question 1

A researcher is concerned that the mean weight of honey bees is more than 11g.

a) H0 : μ = 11 H₁ : μ ≠ 11 b) H0 : μ = 11 H1 : μ > 11 c) H0 : μ = 11 H1 : μ < 800 d) I do not know.

Question 2

The variance in the amount of salt in granola is different from 3.42mg.

a) H0 : σ = 3.42 H1 : σ ≠ 3.42 b) H0 : σ ≠ 3.42 H1 : σ = 3.42 c) H0 : σ² = 3.42 H1 : σ² ≠ 3.42 d) H0 : σ² ≠ 3.42 H1 : σ² = 3.42

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Question 3

3% of homes remain unsold after 6 months in Newark, NJ. It is assumed that homes in the vicinity of the college have a faster sales rate. What null and alternate hypotheses should be used to test this theory?

a) H0 : p = .03 H₁ : p ≠ .03 b) H0 : p = .03 H₁ : p > .03 c) H0 : p = .03 H1 : p < .03 d) H0 : p < .03 H1 : p = .03

Question 4

A company that produces snack food uses a machine to package the product in 454 gram quantities. They wish to test the machine.

Question 5

The NCAA believes 57.6% of football injuries occur during practice. A head trainer thinks this is too high.

Question 6

A manufacturer of machine parts must have a standard deviation in measurement not more than 0.32 mm. Use a sample of 25 parts to test if the standard deviation is outside of the requirement.

F. Statistical Tests

I. Process

• A statistical test compares the sample statistic with the null hypothesized value to decide whether or not the null hypothesis should be rejected.

• The numerical value obtained from a statistical test is called the test value or test statistic.

II. Errors

a) Possible Outcomes of a Hypothesis Test

H0 True H0 False

Reject H0 Error Type I Correct Decision Do not reject H0 Correct Decision Error Type II

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 4 b) Types of Errors

• A Type I error occurs if one rejects the null hypothesis when it is true.

o The maximum probability of committing a type I error:

P (type I error) ≤ ^α

o The probability of rejecting a true null hypothesis.

o Also called the level of significance or α.

o Set by the researcher. If not stated, use 0.05.

• A Type II error occurs if one does not reject the null hypothesis when it is false.

o P (type II error) = β

o The probability of not rejecting a false null hypothesis.

o Decreases when α increases.

c) Level of significance ∝

• Maximum probability of type I error

• P(type I error)≤∝

• The probability of rejecting a true null hypothesis

• Set by researcher

• If not note, use 0.05 d)

In most hypothesis testing situations, β cannot easily be computed.

α (alpha) and β (beta) Probabilities

However, decreasing either alpha or beta increases the other.

II. Hypothesis Testing of a Specified Value for the Population Mean: The Traditional or Critical Value Method

(This is a “comparison of locations method”) A. Terms

I. The rejection region is the range of test values that indicates a significant difference exists between the sample statistic and the hypothesized parameter. The critical value marks the start of the rejection region or critical region. If the test statistic is located in the critical region, the null hypothesis should be rejected.

II. The non-critical or non-rejection region is the range of values of the test value that

indicates that the difference was probably due to chance. If the test value is located in the non-critical region, the null hypothesis should not be rejected.

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 5 III. Critical Value (c.v.)

• Separates the critical region or rejection region from the non-critical region or non- rejection region.

• It is determined from the appropriate table.

• It is based on the significance level and the alternate hypothesis.

B. Finding Critical Values

I. Concepts for One-Tailed Tests

• The critical region or rejection region is only on ONE side of the center value for the distribution or in one tail.

• It is Right-tailed or left-tailed, depending on the direction of the inequality of the alternate hypothesis.

• The area of one tail is equal to the level of significance.

a) Left-Tailed z-tests or t-tests H0 : μ = k H1 : μ < k

• This is a left tailed test since the alternate hypothesis has the “<” symbol.

• Since the alternate hypothesis is less than k, the rejection region is to the left of a negative z critical value.

Note: Whenever the alternative hypothesis is “less than”, the rejection region is to the left of the center of the normal distribution. H1 contains “<” which points to the left  like a left pointing arrow.

0 -z

Critical region

Noncritical region

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• To find a left tailed z critical value (c.v.), locate the level of significance or alpha in the center portion of the Normal Probability Table. Move your hand along the row to the left to find the units and tenths digits for z. Next, move your hand to the top of the column to find the hundredth digits for z. Notice this value is negative.

• To find a left tailed t critical value, use the t table. Find the one tailed α row at the top of the table. Use the column with the specified value of α for the problem. Move your hand down this column to the row for the df in the problem. Affix a negative sign to this value because the cv is less than 0.

•

For a left-tailed test, find critical z for α = 0.01

.

Example:

Since the test is left-tailed, alternate hypothesis contains a “less than” and the rejection region is on the left. Locate 0.0100 inside of the normal probability table. Move along the row to the left to -2.3. Again, starting at 0.0100, move to the top of the column to .03. The critical value is – ( 2.3 + .03). The rule would be to reject the null hypothesis if the test value is less than or equal to -2.33.

b) Right-Tailed z-tests or t-tests H₀ : μ = k H₁ : μ > k

• This is a right tailed test since the alternate hypothesis has the “>” symbol.

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• Since the alternate hypothesis is more than k, the rejection region is to the right of a positive c.v.

Note: Whenever the alternative hypothesis is “greater than” or has the “>”

symbol, the rejection region is to the right of the critical value (c.v.) and to the right of the center of the normal distribution. H1 contains > which points to the right like a right pointing arrow .

• To find a one tailed, right tailed z critical value (c.v.), locate (1 – α) or 1 minus the level of significance in the center portion of the Normal Probability Table. Move your hand along the row to the left to find the units and tenths digits for z and move your hand to the top of the column to find the hundredth digits for z.

Notice this c.v. is positive.

• To find a one tailed, right tailed t critical value, use the t-table. Use the one tailed row at the top of the page. Find the column for t and move your hand down this column until you reach the row for n – 1 degrees of freedom. The entry at the intersection of column t and row df is the c.v. DO NOT MAKE IT NEGATIVE.

0 +z

Critical region Noncritical

region

𝛼 1 − 𝛼

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0 ^1.65

•

Use the z table to find the critical value for a right-tailed test for α = 0.05.

Example:

Since the test is right-tailed, the alternate hypothesis contains the “more than”

symbol and, the rejection region is on the right. Locate 1 – .05 = .9500 inside the normal probability table. Since it is exactly half way between 1.64 and 1.65, the critical value would be 1.645 or rounded to two decimal places it would be 1.65.

Thus reject for any test value is greater than or equal to 1.65.

II. Concepts for Two-Tailed Tests

• The alternative hypothesis always has H0 : μ = k H1 : μ k

• Since the alternative hypothesis has , the test statistic can be either on the right or on the left of the center.

• The null hypothesis is rejected when the test value is in either of 2 critical or rejection regions (one on the left and one on the right).

• It is necessary to find critical values (c.v.) for both sides.

• The critical values are opposites (only f the z and t tables or cv = ± c value).

• The sum of the areas in the two tails is equal to the level of significance or alpha.

That is, each tail has probability or area

Note: every two-tailed test has the sign, 2 critical values, and 2 rejection regions.

Each tail has probability or area .

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•

Using the z-table, find the critical values for a two tailed test when α = 0.01.

Example:

For a two-tailed test, the alternative hypothesis will be “not equal to” and the rejection region is on the right and left with each area 0.01/2 = 0.005. To find the critical value on the left, locate 0.005 inside the normal probability table. The left critical value is -2.575 or rounded to two decimal places is -2.58. Using symmetry, the critical value one the right is 2.58. The rule would be to reject the null hypothesis for any text value greater than or equal to 2.58 AND also to reject the null hypothesis for any test value smaller than -2.58. CV = ± 2.58

Question 7

Using the z-table, find the critical value for a right tailed test with = .025.

Question 8

Use the z-table, find the critical value for a left tailed test with = .10.

Question 9

Use the z-table, find the critical values for a two tailed test with = .05.

III. Combining Concepts

•

In 2002, the mean age of an inmate on death row was 40.7 years, according to data obtained from the U.S. Department of Justice. A sociologist wants to test the statement using the 0.10 level of significance. State the null and alternate hypotheses. State the critical value or values, and the rejection rule.

Example:

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 10 Since the sociologist is not implying a direction, there is none and the hypotheses are:

H0 : μ = 40.7 H1 : μ ≠ 40.7

Since the alternative hypothesis is ≠, the test is two tailed with 0.10/2 = 0.05 probability in each tail. Use the Normal Probability table backwards to find the z value with 0.05 in the left tail. The left critically value is z = -1.65. By symmetry the right critical value is the opposite or +1.65. The rule is to reject the null hypothesis if the test statistic is either greater than 1.65 or less than -1.65. CV = ± 1.65.

Question 10

An energy official thinks that the oil output per well has declined from the 1998 level of 11.1 barrels per day. Use the .001 level of significance. State the null and alternative

hypotheses. State the critical value or values, and the rejection rule.

C. Hypothesis Testing Process using the Critical Value or Traditional Method (location method)

1. State the hypothesis.

2. Compute the test value or test statistic.

3. Find the critical value or beginning of the rejection region or regions from the appropriate table.

4. Decide to reject or fail to reject the null hypothesis based on the location of the test value.

5. Record a conclusion in terms of the situation in the problem.

Note: This is very, very important – MEMORIZE THIS!!!

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 11 I. Test Statistics (to test the value of the Population Mean)

1. z Test for the Population Mean

Use z when 𝜎, the population standard deviation, is know and either n ≥ 30 or the population is normally distributed.

𝑍 = _𝜎/√𝑛^{𝑋�−𝜇} 𝑧 = (𝑋� − 𝜇) ÷ �𝜎 ÷ �(𝑛)�

Where:

𝑋�= sample mean

𝜇= hypothesized population mean 𝜎= population standard deviation 𝑛= sample size

2. t Test for the Population Mean

Use the t Test when 𝜎 is not known and either n ≥ 30 or the population is normally distributed.

𝑡 =_𝑠/√𝑛^{𝑋�−𝜇} 𝑑𝑓: 𝑛 − 1 𝑡 = (𝑋� − 𝜇) ÷ (𝑠 ÷ �(𝑛))

Your calculator work will be the same for every problem testing the specified value for the mean.

II.

•

Problems:

According to the USA Today, the average age of commercial jets in the U.S. is 14 years. An executive of a large airline selects a sample of 36 planes. The average age of the jets in this sample is 11.8 years. The population standard deviation is 2.7 years. At the 0.01 level of significance, is the data sufficient to conclude that the average age of the planes in this company is less than the national average?

Example:

1. State the null and alternate hypotheses H0 : μ = 14 H1 : μ < 14

2. Find the test statistic 𝑧 =^11.8−14_�2.7

√36� = (11.8 − 14) ÷ �2.7 ÷ √36� = −4.89

*Use z since the population standard deviation is stated and n = 36 ≥ 30.

NOTICE THAT THE TEST VALUE IS COMPUTED USING SAMPLE DATA.

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 12 3. Find the critical value. This is a one tailed test with a level of significance equal to 0.01 or the probability in the left tail is 0.01. Since this is a z test statistic, use the normal probability table to find: c.v. = -2.33

NOTICE THAT THE REJECTION REGION IS DEFINED BY A TABLED CRITICAL VALUE.

4. Since -4.89 is less than -2.33 or the test value is in the rejection region, reject the null hypothesis.

5. The data supports an average age of the planes in the executive’s airline to be less than the national average of 14 years.

•

Two researchers measured the pH of randomly selected lakes in the Southern Alps:

Example:

7.2 7.3 6.1 6.9 6.6 7.3 6.3 5.5 6.3 6.5 5.7 6.9 6.7 7.9 5.8

It is assumed that pH is a normally distributed random variable. Based on this sample data, can we conclude that on average, lakes in the Southern Alps are non-acidic or have a pH higher than 6.0?

Use your calculator to find the sample average and sample standard deviation. Since the researcher used a random samples of lakes, use 𝑋� and 𝑠 from the calculator.

𝑋� = 6.60 𝑠 = .672 𝑛 = 15

1. Since there is a concern about a pH “higher than 6.0” use:

H0 : μ = 6.0 H1 : μ > 6.0

2. Use the t test statistic since the population standard deviation is not given but the variable is normally distributed with n = 15 < 30.

𝑡 =^(6.60−6.0)_�.672

√15� = (6.60 − 6.00) ÷ �. 672 ÷ �(15)� = 3.46

NOTICE THAT THE TEST VALUE IS COMPUTED USING SAMPLE DATA.

3. Use the t-table with df: 15 – 1 = 14 to find c.v. = 1.761. The rejection region is to the right of 1.761.

NOTICE THAT THE REJECTION REGION IS DEFINED BY A TABLED CRITICAL VALUE.

4. Since the test value 3.46 is larger than the c.v. = 1.761, it is in the rejection region. Reject the null hypothesis.

5. Since the null hypothesis is rejected, it is removed and the hypothesis that remains is the alternate hypothesis. The data supports an average pH greater than 6.0 for lakes in the Southern Alps.

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Question 11

Concerned that adult females under 51 years are not getting adequate iron intake, a

statistician selected a random sample of women under 51 years old and found the listed iron intake in milligrams for a 24-hour period. Use a 5% level of significance and test his concern.

Assume RDA for iron is 18mg and assume it is a normally distributed random variable.

15.0 18.1 14.4 14.6 10.9 18.2 18.3 15.0 11.0 15.6 11.6 19.8 20.7 13.1 12.1 11.5

Question 12

To see if young men ages 8 through 17 years spend an average of $24.44 per shopping trip to a local mall, the manager surveyed 33 young men and found the average amount spent per visit was $22.97. The standard deviation of the sample was $3.70. Assume the amount spent on a shopping trip is normally distributed. At α = 0.02, can it be concluded that the average amount spent at a local mall is not equal to the national average of $24.44?

III. Hypothesis Testing: P – value, prob – value or comparison of areas method A. Computation of P

• For a one-tailed test: the P-value is the area from the test statistic to more extreme values in the direction of the alternative hypothesis.

• For a two-tailed test: the P-value is twice the area from the test statistic to the end of the tail. If the test value is less than zero, the area is to the left. If the test value is more than zero, the area is to the right.

B. Process for the P-value Method (comparison of areas method)

1. State the hypotheses.

2. Compute the test value.

3. Find the P-value or area in the tail or tails past the test statistic.

Notice the test statistic is used to find the P-value.

4. Decision Rule: reject the null hypothesis whenever p is less than or equal to the level of significance.

• If P-value ≤∝, reject the null hypothesis

• If P-value >∝, fail to reject the null hypothesis NOTE: This is very, very important – MEMORIZE THIS!!!

5. Record a conclusion related to the problem.

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C. TI-83 Directions for Normal and T Probability

(Use the 2^nd function VARS to access DISTR)

1. Use 2^nd function VARS to enter the DISTR Menu 2. Select normalcdf (for z) or tcdf (for t)

3. For z, enter the values for the left end point and the right endpoint (separated by commas) for the area or probability to be determined

• normalcdf (left endpoint, right endpoint)

• For H1 : μ < μ0, p = normalcdf (-100, test value)

• For H1 : μ > μ0, p = normal cdf (test value, 100)

• For H1 : μ ≠ μ₀, p = 2normal cdf (|𝐭𝐞𝐬𝐭 𝐯𝐚𝐥𝐮𝐞|, 100)

4. For t, enter the values for the left and right endpoints and degrees of freedom for t.

• tcdf (left endpoint, right endpoint, df)

• For H1 : μ < μ0, p = tcdf (-100, test value, df)

• For H1 : μ > μ₀, p = tcdf (test value, 100, df)

• For H1 : μ ≠ μ₀, p = tcdf (|𝐭𝐞𝐬𝐭 𝐯𝐚𝐥𝐮𝐞|, 100, df)

Refer to the Chapter 6 Guide for more information on the normalcdf.

The average stopping distance of a school bus traveling 50 miles per hour is usually 26 feet (Snapshot, USA TODAY, March 12, 1992). A group of automotive engineers

determined the average stopping distance for 30 randomly selected busses, traveling 50 miles per hour to be 262.3 feet. The standard deviation of the population was 3 feet.

Use this data to test if the average stopping distance is actually less than 264 feet.

Example:

Use the P-value method with α = 0.01.

1. H0 : μ = 264 H1 : μ < 264 2. 𝒛 = ^𝑿−𝝁^𝝈

√𝒏

= ^{𝟐𝟔𝟐.𝟑−𝟐𝟔𝟒}𝟑

√𝟑𝟎

= (𝟐𝟔𝟐. 𝟑 − 𝟐𝟔𝟒) ÷ �𝟑 ÷ √𝟑𝟎� = −𝟑. 𝟏𝟎

*Use z since the population standard deviation is stated and n = 36 ≥ 30.*

3. Since this is a one tailed test, find the area less than -3.10 or P(z < -3.10) = .0010.

The P-value is .0010 = normalcdf (-100, -3.10) 4. Reject the null hypothesis since P = .0010 < α = .010.

5. The data supports an average stopping distance less than 264 feet.

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 15 Example:

The average salary of graduates entering the actuarial field is reported to be $500000.

To test this figure, a statistics professor surveyed 20 graduates. Their average salary is

$53,228 with a standard deviation of $4000. Assume that these salaries are normally distributed and use the critical value method with a 0.05 level of significance to test the reported $55000.

Critical Value Method

$53, 228

X = s = $4,000 n = 20 1. H0 : μ = 50,000 H1 : μ ≠ $50,000 2. 𝒕 = ^𝑿−𝝁^𝒔

√𝒏 = 𝟓𝟑,𝟐𝟐𝟖−𝟓𝟎,𝟎𝟎𝟎 𝟒𝟎𝟎𝟎

√𝟐𝟎

= −𝟑. 𝟔𝟏 3. c.v. = ± 2.093 d.f. = 19

4. The test statistic 3.61 is further out in the tail past the cv = 2.093. Reject the null hypothesis.

5. The data is sufficient to reject or refute an average salary equal to $50,000.

The average salary of graduates entering the actuarial field is reported to be $500000.

To test this figure, a statistics professor surveyed 20 graduates. Their average salary is

$53,228 with a standard deviation of $4000. Assume that these salaries are normally distributed and use P-value method with a 0.05 level of significance to test the reported

$55000.

Example: P – Value Method

1. H0 : μ = 50,000 H1 : μ ≠ $50,000 2. 𝑡 = ^𝑋−𝜇^𝑠

√𝑛 = 53,228−50,000 4000

√20

= −3.61

3. For a two tailed test, be sure to double the tail area past the test statistic.

4. Find twice the area in the tail from the test value -3.61 to the end of the left tail or P

= 2P(t < -3.61) = 2tcdf(-100, -3,61, 19) = .0019

⁻^2.093

₀

^2.093

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 16 5. Since P is less than the level of significance 0.05, reject the null hypothesis.

6. The data is sufficient to reject an average salary equal to $50,000.

In 2001, the mean household expenditure for energy was $1493, according to data obtained from the U.S. Energy Information Administration. An economist wants to know whether this amount has changed significantly from its 2001 level. Using a random sample of 35 household, she finds the mean expenditure (in 2001 dollars) for energy during the most recent year to be $1618, with standard deviation $321. Complete the test for the economist at the 0.05 level of significance using the P-value method.

Example:

.

𝑿 = 𝟏𝟔𝟏𝟖, 𝒔 = 𝟑𝟐𝟏, 𝒏 = 𝟑𝟓 1. H0: μ = 1493 H1: μ ≠ 1493

2. Use t since the standard deviation for the population is not given and n=35 >30 𝑡 = 𝑋 − 𝜇

𝑠

√𝑛

= 1618 − 1493 321

√35

= (1618 − 1493) ÷ �321 ÷ √35� = 2.30 3. This is a two tailed test since the alternate hypothesis has ≠.

The degrees of freedom are n – 1 or 35 -1 = 34

Calculator: P = 2P(t >2.30) =2tcdf (2.30,100, 34) = .0277

Table: Using the row for degrees of freedom 34, 2.30 is between the values 2.032 (go to the top of its column) with two tailed probability 0.05 AND 2.441 (go to the top of its column) with two tailed probability 0.02. This means P is bounded by 0.05 and 0.02 or 0.02 < P < 0.05

4. From the calculator, P = .0277 and P < .05. Reject the null hypothesis.

From the table, since the upper bound for P is 0.05, P < .05. Reject the null hypothesis.

5. The data is sufficient to reject an average energy expenditure equal to $1493.

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Question 13

Last year the average cost of a concert ticket was $54.80. This year, a random sample of 15 recent concerts had an average price of $62.30 with a variance of $90.25. At the 0.05 level of significance, can it be concluded that the cost has increased?

Question 14

A study published in the American Journal of Psychiatry measured the effect of alcohol on the developing hippocampus, or the portion of the brain responsible for long term memory.

To determine if the volume of the hippocampus is less than the normal 9.02 cm³for adolescents who abuse alcohol, the research used a sample of 12 adolescents with alcohol abuse problems. The average weight of their hippocampus was 8.10 cm³ with a standard deviation of 0.7cm³. Use the 0.01 level of significance.

Question 15:

The U.S. golf Association requires that golf balls have a diameter that is 1.68 inches. An engineer for the USGA wishes to discover whether Maxfli XS golf balls have a mean

diameter different from 1.68 inches. A random sample of Maxfli Xs golf balls was selected.

Assume the diameters are normally distributed and test with a 0.10 type one error rate.

Conduct the test using the P-value method.

1.683 1.677 1.681 1.685 1.678 1.686 1.684 1.684 1.673 1.685 1.682 1.674

Using a calculator find: 𝑋 = 1.6810, 𝑠 = 0.0045, 𝑛 = 12 and then conduct the test.

Question 16:

A Gallup Poll stated that women visit their physician an average of 5.8 times a year. The number of physician visits for 2007 is listed below for 20 randomly selected women:

3 2 1 3 7 2 9 4 6 6 8 0 5 6 4 2 1 3 4 1

Assume that the number of physician visits is normally distributed. At the 0.05 level of significance can we conclude that the Gallup Poll average is correct? Use the p-value method.

𝜇 = 5.8 Use a calculator to find: 𝑋� = 3.85 n = 20 s = 2.52 ∝ =0.05

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IV.Hypothesis Tests for a Proportion (Always use z) A. Formula

Use this method only when 𝒏𝒑 ≥ 𝟓 and 𝒏(𝟏 − 𝒑) ≥ 𝟓 𝒛 = ^{𝒑�− 𝒑}

�^𝒑𝒒_𝒏 = (𝒑� − 𝒑) ÷ �(𝒑 ∙ 𝒒 ÷ 𝒏) where

𝒑� =^𝑿_𝒏

𝒑 = 𝒉𝒚𝒑𝒐𝒕𝒉𝒆𝒔𝒊𝒛𝒆𝒅 𝐩𝐫𝐨𝐩𝐨𝐫𝐭𝐢𝐨𝐧 𝒏 = 𝐬𝐚𝐦𝐩𝐥𝐞 𝐬𝐢𝐳𝐞

Note: the positions of the sample proportion 𝒑� and the hypothesized proportion p in the formula.

B. Problem Characteristics:

• all involve a hypothesis about a percent, proportion, or fraction

• change all percents, proportions, or fractions to four place decimals before computing the test statistic

• If necessary, form 𝑝̂ =^𝑋_𝑛 where x is the count of the number of success and n is the maximum number of tries

• 𝑝̂ =^𝑋_𝑛 is derived from data

•

In a survey conducted by the American Animal Hospital Association, 37% of the

respondents state that they talk to their pets on the answering machine or telephone. A veterinarian thought this percent was high. He randomly selected 150 pet owners. 54 of them responded that they speak to their pet on the answering machine or telephone.

Use this data to run the test at the 0.10 level using the p-value method.

Example:

P = 0.37 𝒑� =_𝟏𝟓𝟎^𝟓𝟒 = 𝟎. 𝟑𝟔 is derived from data 1. H0: p = 0.37 H1: p < 0.37 (claim)

2. The test statistic is:

𝑧 = ^{.36− .37}

�(.37)( .63) 150

= (.36 − .37) ÷ �(. 37 ∙ .63 ÷ 150) = -.26 3. Since this is a one tailed test, find the area less than - 0.26 or

P = P(z < -0.26) =.3974 or

P = normalcdf(-100, -.26) = .3974

4. Fail to reject the null hypothesis since P = .3974 > α = .010.

5. The data does not support less than 37% of pet owners talking to their pets on the telephone or answering machine.

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Question 17:

It has been reported that 40% of the adult population over 60 use e-mail. From a random sample of 180 adults, 65 used e-mail. At ∝ = 0.01, is there sufficient evidence to conclude that the proportion differs from 40%? Use the P-value method.

Question 18:

USA TODAY reported that 63% of Americans will take a vacation this summer. In a survey of 143 Americans 85 were planning to vacation this summer. Use this data to test the USA TODAY report at the 0.05 level of significance with the critical value method.

V. Hypothesis Tests for One Variance (always use the Chi-Squre Test) A. Formula

With d.f. = n – 1 where

Notice the position of the sample variance and the hypothesized variance in this formula.

B. Review of Chi-Square Distributions

The chi-square distributions are a family of probability distributions identified by degrees of freedom (n-1) with:

1. Zero or positive values

2. Distribution skewed to the right

3. One mode slightly to the left of the degrees of freedom Important: The mode slightly left of degrees of freedom

Chi-Square Distributions depend on degrees of freedom

Note: as df increases the curve moves to the right

If degrees of freedom are not listed in the table, use the closest smaller value.

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C. Comments

• Also used to test a value of a standard deviation

(square the sample and hypothesized standard deviation to find the variances)

• Can be left tailed, right tailed, or two tailed

• Left tailed test statistics should be to the left of df

• Right tailed test statistics should be to the right o df

• Two tailed test statistics can be either on the left or the right of the df

• Mark the df slightly to the right of the mode. Now you know the approximate location of the center.

• The value of the c.v. is always positive for 𝑥²

•

A researcher believes that the standard deviation of the number of cars stolen each year is less than 15. For a sample of 12 years, the standard deviation for the number of stolen aircrafts is 13.6. Use ∝ = 0.05 and the critical value method for the test.

Example:

Since the “researcher believes that the standard deviation of the number of cars stolen each year is less than 15”, the test is a left tailed test. The test statistics should be on the left of the df and the critical value will be to the left of the degrees of freedom which is approximately at the mode. The p-value will be the area from the test statistic to the left end of the tail.

1. H0 : 𝝈 = 15 H1 : 𝝈 < 15 2. 𝒙^𝟐 =^{(𝒏−𝟏)𝒔}_𝝈_𝟐 ^𝟐 = (𝟏𝟐−𝟏)∙𝟏𝟑.𝟔^𝟐

𝟏𝟓^𝟐 = 𝟏𝟏 ∙ 𝟏𝟑. 𝟔^𝟐÷ 𝟏𝟓^𝟐= 𝟗. 𝟎𝟒

*Since this is a left tailed test, use the 1- ∝ = 1 - .05 = .95 𝒙^𝟐 column. With df: n – 1

= 12 – 1 = 11 to find c.v.

3. C.V. = 4.575 d.f. = 11

Note: This c.v. is always positive.

4.575 11

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 21 4. The test statistic 9.04 is not less than the critical value 4.575. Do not reject the null

hypothesis.

5. The data is not sufficient to reject a standard deviation equal to 15.

D. TI-83 Calculator: P-values for the Chi-Square Statistics

• For left tailed tests, the test statistic is left of df and p is the area from zero to the test statistics or the area on the left or

• p = (0, test statistic, df)

• For right tailed tests, the test statistic is right of df and p is the area from the test value to the end of the right tail or

• p = (test statistic, 1000, df)

• For two tailed tests, first determine if the test value is to the right or left of the df

• If the test value is less than df, p = 2 (0, test statistic, df)

• If the test value is greater than df, p = 2 (test statistic, 1000, df)

•

A random sample of 20 different kinds of doughnuts had the calories listed above. At ∝

= 0.01, is there sufficient evidence to conclude that the standard deviation of calorie content is greater than 20 calories? Use the critical value method.

Example:

290 320 260 220 300 310 310 270 250 230 270 260 310 200 250 250 270 210 260 300 Use your calculator to find the sample standard deviation. s = 35.11 1. H0 : H1 :

2.

3. The c.v. is 36.191.

4. For a right tailed test, the cv is to the right of degrees of freedom. Since the test statistic is further to the right than the cv, reject the null hypothesis.

5. The data is sufficient to support a standard deviation greater than 20.

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Question 19:

The manager of a large company is concerned about the variability of the time that it takes a telephone call to be transferred to the correct office in her company. A sample of 15 calls is selected, and the transfer time is recorded. The standard deviation of the sample is 1.8 minutes. At ∝ = 0.01, test if the population standard deviation is more than 1.2 minutes.

Use the P-value method.

Question 20:

The data below is a random sample of home run totals for National League Champions from 1938 to 2001. At the 0.05 level of significance, is there sufficient evidence to conclude that the variance is smaller than 70? Use the prob-value method and the cv method.

34 47 43 23 36 50 42

44 43 40 39 41 47 45

VI.Summary

• A statistical hypothesis is a conjecture about a population mean, proportion, or variance.

• There are two hypotheses: the null hypothesis states that there is no difference (=) and the alternative hypothesis specifies a difference ( <, >, or ≠).

• Researchers compute a test value from the sample data to decide whether the null hypothesis should or should not be rejected.

• If null hypothesis is rejected, the difference between the population parameter and the sample statistic is said to be significant.

• The difference is determined to be significant when either:

• the test value falls in the critical region or

• the p-value is less than or equal to α, the level of significance of the test.

• The significance level of a test is the probability of committing a type I error.

• The significance level is usually specified in the problem. The default value for the level of significance is 0.05.

• A type I error occurs when the null hypothesis is rejected when it is true.

• The type II error can occur when the null hypothesis is not rejected when it is false.

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Answer: Question 1

A researcher is concerned that the mean weight of honey bees is more than 11g.

a) H0 : μ = 11 H1 : μ ≠ 11 b) H0 : μ =11 H1 : μ > 11 c) H0 : μ = 11 H₁ : μ < 800 d) I do not know.

The variance in the amount of salt in granola is different from 3.42mg.

a) H0 : σ = 3.42 H₁ : σ ≠ 3.42 b) H0 : σ ≠ 3.42 H₁ : σ = 3.42 c) H⁰: σ² = 3.42 H¹: σ² ≠ 3.42 d) H0 : σ² ≠ 3.42 H1 : σ² = 3.42

3% of homes remain unsold after 6 months in Newark, NJ. It is assumed that homes in the vicinity of the college have a faster sales rate. What null and alternate hypotheses should be used to test this theory?

a) H0 : p = .03 H1 : p ≠ .03 b) H0 : p = .03 H₁ : p > .03 c) H⁰ : p = .03 H¹ : p < .03 d) H0 : p < .03 H1 : p = .03

A company that produces snack food uses a machine to package the product in 454 gram quantities. They wish to test the machine.

H0 : μ = 454 H1 : μ ≠ 454

The NCAA believes 57.6% of football injuries occur during practice. A head trainer thinks this is too high.

H0 : p = .576 H1 : p < .576

A manufacturer of machine parts must have a standard deviation in measurement not more than 0.32 mm. Use a sample of 25 parts to test if the standard deviation is outside of the requirement.

H0: σ = 0.32 H1: σ > 0.32

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Use the z-table, find the critical value for a right tailed test with α = .025.

Since this is a right tailed test, the alternate hypothesis has a “greater than” symbol and the rejection region is on the right. Locate 1 - .025 = .0750 inside the normal probability table. Moving to the left and then moving to the top of the table, the z value is 1.96. The c.v. is 1.96 and the rule would be to reject the null hypothesis if the test value is greater than 1.96.

Use the z-table, find the critical value for a left tailed test with α =.10.

Since the test is left-tailed, alternate hypothesis contains a “less than” and the rejection region is on the left. Locate 0.100 inside of the normal probability table. Move along the row to the left to - 1.2. Again, starting at 0.100, move to the top of the column to .08. The critical value is – ( 1.2 + .08). The rule would be to reject the null hypothesis if the test value is less than or equal to -1.28.

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Use the z-table, find the critical values for a two tailed test with α = .05.

For a two-tailed test, the alternate hypothesis will be “not equal to” and the rejection region is on the right and left with each area 0.05/2 = 0.025. To find the critical value on the left, locate 0.025 inside the normal probability table. The left critical value is -1.96. Using symmetry, the critical value on the right is +1.96. The rule would be to reject the null hypothesis for any test value greater than or equal to 1.96 AND also to reject the null hypothesis for any test value smaller than -1.96. CV = ± 1.96

An energy official thinks that the oil output per well has declined from the 1998 level of 11.1 barrels per day. Use the .001 level of significance. State the null and alternate hypotheses. State the critical value or values, and the rejection rule.

Since the energy official “thinks that the oil output per well has declined”, the hypotheses are:

H0: μ = 11.1 H1: μ < 11.1

The test is left tailed since the alternate hypothesis has the “<” symbol. The rejection area in the one tail is equal to the level of significance or .001.

Use the Normal Probability table backwards to find the z value with 0.001 in the left tail. There are several z values with left tail probability equal to 0.001. Use the center z value or -3.09. The rule is to reject the null hypothesis if the test statistic is less than -3.09.

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Concerned that adult females under 51 years are not getting adequate iron intake, a statistician selected a random sample of women under 51 years old and found the listed iron intake in milligrams for a 24-hour period. Use a 5% level of significance and test his concern. Assume RDA for iron is 18mg and iron intake is normally distributed.

.

15.0 18.1 14.4 14.6 10.9 18.2 18.3 15.0 11.0 15.6 11.6 19.8 20.7 13.1 12.1 11.5

Use a calculator to find: 𝑋 = 14.99, 𝑠 = 3.22, 𝑛 = 16 1. H0 : μ = 18.0 H1 : μ < 18.0 (claim)

“not getting enough calcium” translates into an alternate hypothesis μ < 18.0

2. Use t since the population standard deviation is not given and the distribution is normal.

𝒕 =(𝟏𝟒. 𝟗𝟗 − 𝟏𝟖. 𝟎𝟎)

�𝟑. 𝟐𝟐

√𝟏𝟔� = (𝟏𝟒. 𝟗𝟗 − 𝟏𝟖. 𝟎𝟎) ÷ �𝟑. 𝟐𝟐 ÷ �(𝟏𝟔)� = −𝟑. 𝟕𝟒

3. Use the t-table with 𝒅𝒇– 𝟏𝟔– 𝟏 = 𝟏𝟓, because the population standard deviation σ is not given, but the variable is normally distributed. c.v. = -1.753. Since this is a left-tailed test, the

rejection region is to the left of -1.753.

4. Since the test value -3.74 is to the left of c.v.=-1.753, it is in the rejection region. Reject the null hypothesis.

5. The data is sufficient to support the average daily iron intake is less than 18 for women under 51 years.

To see if young men ages 8 through 17 years spends an average of $24.44 per shopping trip to a local mall, the manager surveyed 33 young men and found the average amount spent per visit was

$22.97. The standard deviation of the sample was $3.70. Assume that the amount spent is

normally distributed. At α = 0.02, can it be concluded that the average amount spent at a local mall is not equal to the national average of $24.44?

.

1. H0 : μ = $24.44 H1 : μ ≠ $24.44 2. 𝒕 =(𝟐𝟐.𝟗𝟕−𝟐𝟒.𝟒𝟒)

�^{𝟑.𝟕𝟎}_√𝟑𝟑� = (𝟐𝟐. 𝟗𝟕 − 𝟐𝟒. 𝟒𝟒) ÷ �𝟑. 𝟕𝟎 ÷ �(𝟑𝟑)� = −𝟐. 𝟐𝟖 3. c.v. = ± 2.449 df: 32.

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 27 4. Do not reject the null hypothesis. -2.28 is the test statistic and is not less than -2.449, nor is it

greater than +2.449.

5. The data is not sufficient to reject a mean average spending per shopping trip for men 8 through 17 years equal to $24.44.

Last year the average cost of a concert ticket was $54.80. This year, a random sample of 15 recent concerts had an average price of $62.30 with a variance of $90.25. Assume the price of concert tickets is normally distributed. At the 0.05 level of significance, can it be concluded that the cost has increased? Use the critical value and the P-value method.

n = 15 𝑋�=$62.30 𝑠² = $90.25 Critical Value Method:

1. H0 : μ = $54.80 H1 : μ > $54.80 (claim) 2.

3. Use the t table to find the critical value with 0.05 in the one tail column and df 14 c.v. = 1.761

4. Since the test statistic (3.06) is further out in the tail past the cv, reject the null hypothesis.

5. The data supports an increase in the average cost of concert tickets.

-2.449 0 2.449

1.761 3.06

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 28 P-value Method:

1. H0 : μ = $54.80 H1 : μ > $54.80 (claim) 2. 𝑡 = ^𝑋−𝜇^𝑠

√𝑛

= ^62.3−54.8_√90.5

√15

= 3.06

3. Find the area in the tail to the right of 3.06 (the test statistic) with df 15 – 1 = 14. The closest value on the chart is 2.997. Since the area in the tail to the right of 2.997 is .005, the area in the tail to the right of 3.06 would have to be less than .005 OR p < .005 using the formula p=tcdf (3.00, 100, 14)

4. Since P < .005, it is less than the level of significance which is .05. Reject the null hypothesis since p is less than alpha.

5. The data supports an average price greater than $54.80.

A study published in the American Journal of Psychiatry measured the effect of alcohol on the developing hippocampus, or the portion of the brain responsible for long term memory. To determine if the volume of the hippocampus is less than the normal 9.02 cm³for adolescents who abuse alcohol, the research used a sample of 12 adolescents with alcohol abuse problems. The average weight of their hippocampus was 8.10 cm³ with a standard deviation of 0.7cm³. Use the 0.01 level of significance.

1. H0 : μ = 9.02 H₁ : μ < 9.02 (claim) 2. 𝒕 = ^𝑿−𝝁^𝒔

√𝒏 = ^{𝟖.𝟏𝟎−𝟗.𝟎𝟐}𝟎.𝟕

√𝟏𝟐

= (𝟖. 𝟏𝟎 − 𝟗. 𝟎𝟐) ÷ �𝟎. 𝟕 ÷ √𝟏𝟐� = −𝟒. 𝟓𝟓

*Use t since the population standard deviation is not stated and n = 12 *

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Dr. Janet Winter, jmw11@psu.edu Stat 200 Page 29 3. Use the row with df: 12 – 1 to find the area less than -4.55. -4.55 is not on the table. -3.055 is

the number closes to it.

P = P( t < -4.55) < P(t < -3.055) = .005 So, the P-value less than .005.

4. Reject the null hypothesis since P < .005 < α = .01.

5. The data supports a decrease in the size of the long term memory portion of the brain for adolescent who abuse alcohol.

The U.S. golf Association requires that golf balls have a diameter that is 1.68 inches. An engineer for the USGA wishes to discover whether Maxfli XS golf balls have a mean diameter different from 1.68 inches. A random sample of Maxfli Xs golf balls was selected. Assume the diameters are normally distributed and test with a 0.10 type one error rate. Conduct the test using the P-value method.

1.683 1.677 1.681 1.685 1.678 1.686 1.684 1.684 1.673 1.685 1.682 1.674 Using a calculator find: 𝑋 = 1.6810, 𝑠 = 0.0045, 𝑛 = 12 1. H0: μ = 1.68 (claim) H1: μ ≠ 1.68

2. 𝑡 = ^𝑋−𝜇^𝑠

√𝑛 = 1.6810−1.68 .0045

√12

= (1.6810 − 1.68) ÷ �0.0045 ÷ √12� = 0.77

*Use t since the population standard deviation is not stated, but the diameters are normally distributed with n = 12.

3. Since this is a two tailed test, find twice the area greater than 0.87 or P = 2P(t > 0.77) = 2tcdf(

0.77, 1000, 11) = .4575

4. Fail to reject the null hypothesis since P = .4575 > α = .010.

5. The data is not sufficient to reject an average diameter equal to 1.68.

Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion

Chapter 8: Hypothesis Testing for One Population Mean, Variance, and Proportion

0 1.65

0

0 ^1.65

₀