Confidence Interval for a Population. Normal (z) Statistic

Confidence Interval for a Population Mean:

Mean:

Normal (z) Statistic

Estimation Process

Mean, µµµµ^{, is} unknown

Population

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I am 95% confident that µµµµis between 40 &

Mean 60.

x = 50

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Sample ☺☺

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Confidence Interval

According to the Central Limit Theorem, the sampling

distribution of the sample mean is approximately normal for large samples. Let us calculate the interval estimator:

x ± 1.96

σ

σ

x ± 1.96

σ

σ

n

That is, we form an interval from 1.96 standard

deviations below the sample mean to 1.96 standard

deviations above the mean. Prior to drawing the sample, what are the chances that this interval will enclose µ,

the population mean?

Confidence Interval

If sample measurements yield a value of that falls between the two lines on either side of µ, then the interval will contain µ.

The area under the

normal curve between

x x ± 1.96

σ

normal curve between these two boundaries is exactly .95. Thus, the probability that a

randomly selected

interval will contain µ is equal to .95.

95% Confidence Level

If our confidence level is 95%, then in the long run, 95% of our confidence intervals will contain µ and 5% will not.

For a confidence level of 95%, the area in the two tails is .05. To choose a different confidence level we increase or decrease the area (call it α) assigned to the tails. If we place decrease the area (call it α) assigned to the tails. If we place α/2 in each tail

and z_α/2 is the z-value, the confidence interval with coefficient (1 – α^{) is}

x ± z

( )

^σ

Conditions Required for a Valid Large- Sample

Confidence Interval for µ

1. A random sample is selected from the target population.

population.

2. The sample size n is large (i.e., n ≥ 30). Due to the Central Limit Theorem, this condition guarantees that the sampling distribution of is approximately normal. Also, for large n, s will be a good estimator of

σ

x

Large-Sample (1 – α α α α )% Confidence Interval for µ

where z_α/2 is the z-value with an area

α

σ

the standard deviation of the sampled population, and n x ± z

( )

^σ

^σ

n

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the standard deviation of the sampled population, and n is the sample size.

Note: When

σ

where s is the sample standard deviation.

x ± z_{α 2} s n

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Thinking Challenge

You’re a Q/C inspector for Gallo.

The σσσσ for 2-liter bottles is .05 liters. A random sample of 100 bottles showed x = 1.99 liters.

What is the 90% confidence interval estimate of the true

mean amount in 2-liter bottles?

2 liter

© 1984-1994 T/Maker Co.

2 liter

Confidence Interval Solution*

x − z_{α/ 2} ⋅ σ n

≤ µ ≤ x + z_{α/ 2} ⋅ σ n

1.99 − 1.645⋅ .05

≤ µ ≤ 1.99 + 1.645⋅ .05

1.99 − 1.645⋅ .05 100

≤ µ ≤ 1.99 + 1.645⋅ .05 100 1.982 ≤ µ ≤ 1.998

We can be 90% confident that the mean amount in 2-liter bottles

between 1.982 and 1.998. Our confidence is derived from the fact that 90% of the intervals formed in repeated applications of this procedure would contain µ

Exercise

• A random sample of 70 observations from a normally distributed population possesses a sample mean equal to 26.2 and a sample standard deviation equal to 4.1

• A) Find an approximate 95% confidence interval for µ.

• B) What do you mean when you say that a confidence

• B) What do you mean when you say that a confidence level is 95%?

• C) Find an approximate 99% confidence interval for µ.

• D) What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed?

Confidence Interval for a Population Mean:

Mean:

Student’s t-Statistic

Small sample size problem for inference about µ

• The use of a small sample in making inference about µ presents two problems when we

attempt to use the standard normal z as a test statistic.

statistic.

Problem 1

• The shape of the sampling distribution of the sample mean now depends on the shape of the population sampled.

• We can no longer assume that the samplingWe can no longer assume that the sampling distribution of sample mean is approximately

normal because the central limit theorem ensures normality only for samples that are sufficiently large.

Solution to Problem 1

• We know that if our sample comes from a population with normal distribution the sampling distribution of sample mean will be normal regardless of the sample size.

Problem 2

• The population standard deviation σ is almost always unknown. For small samples the sample standard deviaiton s provides poor approximation for σ.

Solution to Problem 2

(Small Sample with σ σσ σ ^Unknown)

Instead of using the standard normal statistic z = x − µ

σ

x − µ

σ

x

t = x − µ s n

in which the sample standard deviation, s, replaces the population standard deviation,

σ

Conditions Required for a Valid Small- Sample Confidence Interval for µ

• A random sample is selected from the target population

• The population has a relative frequency distribution that is approximately normal.

distribution that is approximately normal.

Small Sample with σ σσ σ ^known

Use the standard normal statistic z = x − µ

σ

x − µ

σ

x

Student’s t-Statistic

The t-statistic has a sampling distribution very much like that of the z-statistic: mound-shaped, symmetric, with mean 0.

The primary difference between the sampling between the sampling distributions of t and z is that the t-statistic is more variable than the z-statistic.

Degrees of Freedom

The actual amount of variability in the sampling distribution of t depends on the sample size n. A

convenient way of expressing this dependence is to say that the t-statistic has (n – 1) degrees of freedom (df).

that the t-statistic has (n – 1) degrees of freedom (df).

Student’s t Distribution

Standard Normal

t (df = 13) Bell-Shaped

Symmetric

‘Fatter’ Tails

z

0 t

t (df = 5)

‘Fatter’ Tails

The smaller the degrees of freedom for t-statistic, the more variable will be its sampling distribution.

1)

2)

• We have a random sample of 15 cars of the same model. Assume that the gas milage for the population is normally distributed with a standard deviaition of 5.2 miles per galon.

• A) Identify the bounds for a 90% confidence interval for the mean given a sample mean of 22.8 miles per gallon.

• B) The car manufacturer of this particular model claims that the average gas milage is 26 miles per gallon. Discuss the validity of this average gas milage is 26 miles per gallon. Discuss the validity of this claim using the 90% confidence interval calculated in A.

• C) Let a and b represent the lower and upper boundaries of 90%

confidence intervl for the mean of the population. Is it correct to conclude that tere is a 90% probability that true population mean lies between a and b?

Thinking Challenge

• We have a random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.5.

• A) Calculate a 90% confidence interval for the mean given a sample size of 40 orders.

• B) Calculate a 90% confidence interval for the mean

• B) Calculate a 90% confidence interval for the mean given a sample size of 75 orders.

• C) Explain the difference in the 90% confidence intervals calculated in A and B.

• Calculate the minimum sample size needed to identify a 90% confidence interval for the mean assuming a $5 margin of error.