n Using the formula we get a confidence interval of 80±1.64

9.52 The professor of statistics noticed that the marks in his course are normally distributed. He has also noticed that his morning class average is 73% with a standard deviation of 12% on their final exams. His afternoon classes average 77% with a standard deviation of 10%. What is the probability that the mean mark of four randomly selected students from a morning class is greater than the average of four randomly selected students from an afternoon class?

Solution:

X = marks ~ N(µ, σ) Xm~ N(73,12) Xa~ N(77,10) nm=na=4

Since the population standard deviation is known for both morning and afternoon class 

Formula:

)

)

(

(

)

0

(

n

n

X

X

Z

P

X

X

P

σ

σ

µ

µ

+

−

−

−

>

=

>

−

Using this formula we get P(Z>0.51) and reading the value in the Z-table 0.5-0.195=0.31 Answer: The probability that the sample mean in the morning class is larger than the one in the afternoon class is 0.31.

10.11 A random sample of 25 was drawn from a normal distribution whose standard deviation is 5. The sample mean was 80.

a. Determine the 90% confidence interval estimate of the population mean. Solution:

90% confidence interval for the population mean α = 1-0.9=0.1 Formula:

n

Z

X

±

σ

Using the formula we get a confidence interval of 80±1.64 b. Repeat part a. with a sample size of 100.

Answer: 80±0.82

c. Repeat part a. with a sample size of 400. Answer: 80±0.41

d. Describe what happens to the confidence interval estimate when the sample size increases. As the sample size increase we have more information about the population.

10.53 The operations manager of a large production plant would like to estimate the average amount of time workers take to assemble a new electronic component. After observing a number of workers assembling similar devices, she guesses that the standard deviation is 6 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 10 seconds? Assume that the confidence level is to be 90%.

Solution:

σ=6 minutes = 360 seconds

Using the confidence interval formula we can rearrange the formula and get the following formula to calculate n: 2 2 /













⋅

=

W

Z

n

σ

Where W= 10 seconds (W equals the part after the ± in calculation of an CI) Answer: Using the formula we get that the sample size must be of 3486 or more.

4. Introduction to test statistic and p-value

Find the p-value of the following test given that

x

505

:

505

:

>

=

µ

µ

H

H

σ

µ

Using the formula above we get an observed Zobs of -1.77

Use the Z-table to know the p-value for the observed value0.5-0.4616 = 0.038

Answer: The probability to observe a sample mean of 500 or less given that the null hypothesis is true = 0.038

a.

And the p-value is 0.5-0.3810=0.12

Answer: As the population standard deviation increases the p-value increases.

b.

And the p-value is 0.5-0.4938=0.0062

Answer: As the sample size increases the p-value decreases (because we have more information about the population)

c.

x

And the p-value is 0.5-0.1368=0.36

Answer: As the observed sample mean is closer to the null hypothesis population mean the p-value increases.

11.29 A business student claims that on average an MBA student is required to prepare more than five cases per week. To examine the claim, a statistic professor asks a random sample of ten MBA students to report the number of cases they prepare weekly. The results are exhibited here. Can the professor conclude at the 5% significance level that the claim is true, assuming that the number of cases is normally distributed with a standard deviation is 2?

4 12 4 8 9 5 11 3 7 4 Solution: X~N(µ,2) n=10 α=significance level = 0.05 Hypothesis H0: µ =5 HA: µ >5

Calculate the sample mean with formula given in earlier exercises  Sample mean=6.7 Formulate the test statistic (se exercise 4 above).

Rejection region Zobs>Zα=Z0.05=1.645

Then we reject the null hypothesis Observed value 10 / 2 5 7 . 6 − = obs Z = 2.68

Answer: 2.68>1.645 therefore can the professor, on a 5% significance level, conclude that the claim is true.

11.33 An office manager believes that the average amount of time spent by office workers reading and deleting spam e-mail exceeds 25 minutes per day. To test this belief, he takes a random sample of 18 workers and measure the amount of time each spends reading and deleting spam. The results are listed here. If the population of time is normally distributed with a standard deviation of 10 minutes, can the manager infer at the 1% significance level that he is correct?

30 38 19 44 17 21 32 28 34 23 13 9 11 30 42 37 43 48 Solution: X~N(µ,10) n=18 α=significance level = 0.01

H0: µ =25 HA: µ >25

Calculate the sample mean with formula given in earlier exercises  Sample mean=29 Formulate the test statistic (se exercise 4 above).

Rejection region Zobs>Zα=Z0.01=2.33

Then we reject the null hypothesis Observed value 69 . 1 18 / 10 25 29 = − = obs Z

Answer: 1.69 < 2.33 and therefore we cannot reject the null hypothesis.

12.13 A random sample of 8 observations was drawn from a normal population. The sample mean and sample standard deviation are

x

a. Estimate the population mean with 95% confidence.

Since the population standard deviation is unknown and the sample size is small we use the Student t distribution to calculate a 95% confidence interval for the population mean: Formula:

n

s

t

X

±

b. Repeat part a assuming that you know that the population standard deviation is σ = 10

Since we have the population standard deviation we can use the Z-distribution and formula:

n

Z

X

±

σ

This gives an estimated CI of 40±6.93

d.

e.

Using the t-distribution for s=10 and n=100 we will get a CI of 40±1.984 Using the Z-distribution for σ=10 and n=100 we will get a CI of 40±1.96

Because for large sample size (n>30) the Student t distribution is approximately normal distributed. The central limit theorem (CLT).

12.27 Most owners of digital cameras store their pictures on the camera. Some will eventually download these to a computer or print them using their own printers or use a commercial printer. A film-processing company wanted to know how many pictures were stored on cameras. A random sample of 10 digital camera owners produced the data given here. Estimate with 95% confidence the mean number of pictures stored on digital cameras.

25 6 22 26 31 18 13 20 14 2

Solution:

Start by calculating the sample mean and standard deviation with formulas given in earlier practicals/the book or formula sheet 

Sample mean=17.7 s = 9

Since the sample size is small and we don’t know the population standard deviation we use t-distribution to calculate the CI for the population mean=

n

s

t

X

±

17

.

7

±

2

.

262

9

10

12.66 A dean of a business school wanted to know whether the graduates of her school used a statistical inference technique during their first year of employment after graduation. She surveyed 418 graduates and asked about the use of statistical techniques. After tallying up the responses, she found that 217 used statistical inference within 1 year of graduation. Estimate with 90% confidence the proportion of all business school graduates who use their statistical education within a year of graduation.

Solution:

We know that the sample proportion ^p=x/n where x=the number of success (in this case number who used statistics) and n=sample size

^p=0.52

Further we know that the standard deviation for ^p is pˆ(1−pˆ)

To make a 90% confidence interval for the population proportion, p, use the formula:

n

p

p

Z

p

ˆ

±

ˆ

(

1

−

ˆ

)