# Normal Distribution, Sampling distribution

1) If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of sample s = 0.50, the critical t will be:

A) 2.7969 (B) 2.7874 (C) 2.4922 (D) 2.4851

2) A statistician wishes to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Which of the following interpretation is correct?

A) 97% of the sampled total compensation value fell between $2,181,260 and 5,836,180.(B) We are 97% confident that the mean of sampled CEOs fall in the interval $2,181,260 to $5,836,180 (C) In the population of service industry CEOs, 97% of them will have total compensation that fall in the interval $2,181,260 to $5,836,180.(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval $2,181,260 to $ 5,836,180.

3) The personnel of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers revealed the following:

Absenteeism: X = 9.7 days and S=4.0 days

12 clerical workers were absent more than 10days

What are the upper and lower limits of the 95% confidence interval estimate of the mean number of absence for clerical workers last year .

A) 8.132 and 11.268 (B) 8.049 and 11.351 (C) 8.052 and 11.348 (D) 8.331 and 11.069

4) For sample size 16, the sampling distribution of the mean will be approximately normally distributed;

A) regardless of the shape of the population (B) if the shape of population is symmetrical (C) if the sample standard deviation is know (D) if the sample is normally distributed

5) The standard error of the mean for a sample of 100 is 30. In order to manipulate the standard error of the mean to be 15, we would:

A) increase the sample size to 200 (B) increase the sample size to 400 (C) decrease the sample size to 50 (D) decrease the sample to 25

6) A confidence interval was used to estimate the proportion of statistic student that was a female. A random sample of 72 statistics student generated the following 90% confidence interval( 0.438, 0.642) using the information, what size of sample would be necessary if we wanted to estimate the true proportion to within +_ 0.08 using 95% confidence

A) 105 (B) 150 (C) 420 (D) 597

7) A stationery store wants to estimate the total retail value of the 300 greeting cards that it has in its inventory. What are the upper and lower limits of the 95% confidence interval estimate of the population total value of all greeting cards that are inventory if a random sample of 20 greeting card indicates an average value of $1.67 and a standard deviation of $0.32

A) $457.52 and 544.48 (B) $465.08 and $536.92 (C) $460.29 and $541.72 (D) $457.67 and $544.33

8) A major department store chain is interested in estimating the average amount its credit card customer spent on their first visit to the chain's new store in the mall. 15 credits accounts were randomly sampled and analyzed with the following results X=$50.50 s2 and =400.

Assuming the distribution of the amount spent on their first visit is approximately normal, what is the shape of the sampling distribution of the sample mean that will be used to create the desired confidence interval for u?

A) Approximately normal with a mean of $50.50 (B) standard normal distribution (C) A t distribution with 15 degrees of freedom (D) A t distribution with 14 degrees of freedom

9) The standard error of the mean:

A) is never larger than the standard deviation of the population (B) decrease as the sample size increases. (C) measures the variability of the mean from sample to sample (D) All of the above

10) When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimate to be, the _____the sample size required.

A) smaller (B) larger (C) sample size is not effected (D) the effect cannot be determined from the information given

11) Approximately 5% of U.S. families have a net worth in excess of $1 million and thus can be called "millionaires". However , a survey in the year 2000 found that 30% of Microsoft's 31,000 employees were millionaires. If random sample of 100 Microsoft employees had been taken that year, what proportion of the sample would have been between 25% and 35% millionaires ?

A) 0.1379 (B) 0.2758 (C) 0.7242 (D) 1.0911

#### Solution Preview

1) If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of sample s = 0.50, the critical t will be:

A) 2.7969 (B) 2.7874 (C) 2.4922 (D) 2.4851

Answer: A) 2.7969

99% Confidence limits

sample size=n= 25

Confidence level= 99%

Therefore Significance level=α= 1% =100% -99%

No of tails= 2

This is 2 tailed because we are calculating the confidence interval

Since sample size= 25 < 30

and we are using sample standard deviation to estimate the population standard deviation

use t distribution

t at the 0.01 level of significance and 24 degrees of freedom (=n-1) and 2 tailed test= 2.7969

2) A statistician wishes to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Which of the following interpretation is correct?

A) 97% of the sampled total compensation value fell between $2,181,260 and 5,836,180.(B) We are 97% confident that the mean of sampled CEOs fall in the interval $2,181,260 to $5,836,180 (C) In the population of service industry CEOs, 97% of them will have total compensation that fall in the interval $2,181,260 to $5,836,180.(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval $2,181,260 to $ 5,836,180.

Answer:

(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval $2,181,260 to $ 5,836,180.

3) The personnel of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers revealed the following:

Absenteeism: X = 9.7 days and S=4.0 days

12 clerical workers were absent more than 10days

What are the upper and lower limits of the 95% confidence interval estimate of the mean number of absence for clerical workers last year .

A) 8.132 and 11.268 (B) 8.049 and 11.351 (C) 8.052 and 11.348 (D) 8.331 and 11.069

Answer: (B) 8.049 and 11.351

95% Confidence limits

Mean=μ= 9.7 days

Standard deviation =σ= 4.0

sample size=n= 25

σx=standard error of mean=σ/√n= 0.8 = ( 4 /√ 25)

Confidence level= 95%

Therefore Significance level=α= 5% =100% -95%

No of tails= 2

This is a 2 tailed test because we are calculating the confidence interval

Since sample size= 25 < 30

and we are using sample standard deviation to estimate the population standard deviation

use t ...

#### Solution Summary

The solution provides answers and explanations to 11 Multiple choice questions on Normal Distribution, t distribution, confidence interval, sampling distribution of the mean, standard error of the mean etc.