# Statistics - Multiple Choice

Question 1: When a statistic calculated from sample data is used to estimate a population parameter, it is called :

an interval estimate

a point estimate

a statistical parameter

a good guess

Question 2: A large appliance company sends out technicians to unpack, assemble, and connect every gas dryer that is sold. In developing a pricing strategy, it is important to the company to have a "handle" on how long this process takes. The company hires a research assistant to follow technicians on 45 randomly selected jobs. The research assistant records how much time it takes the technicians to unpack, assemble, and connect each gas dryer. The resulting sample mean is 34.3 minutes. If research assistant concludes based on the sample mean that the average time for all such jobs is 34.3 minutes she is using a:

a range estimate

a statistical parameter

an interval estimate

a point estimate

Question 3: The Z value associated with a two sided 90% confidence interval is:

1.28

1.645

1.96

2.575

Question 4: Suppose a random sample of 36 is selected from a population with a standard deviation of 12. If the sample mean is 98, the 99% confidence interval to estimate the population mean is:

94.08 to 101.92

97.35 to 98.65

92.85 to 103.15

93.34 to 102.66

Question 5: What is the average length of a wireless phone call? Suppose researchers in the telecommunications industry want to estimate the average number of minutes of a wireless call. To do so, they randomly select one such call from 85 wireless phone bills around the country. The resulting sample mean is 2.54 minutes with a standard deviation of 1.20 minutes. From these data, the 86% confidence interval to estimate the average length of a wireless phone call for all users is:

2.285 to 2.795

2.326 to 2.754

2.347 to 2.733

2.519 to 2.561

Question 6: The t statistic was developed by:

Charles Student

William Gosset

Abraham deMoivre

Karl Gauss

Question 7: In order to find values in the t distribution table, you must convert the sample size or sizes to:

population sizes

z values

student values

degrees of freedom

Question 8: A researcher is taking a random sample of 18 items in an effort to estimate the population mean. He wants to be 95% confident of his results. The table t value that he should use is:

2.110

2.101

1.740

1.734

Question 9: A researcher is interested in estimating the mean value for a population. She takes a random sample of 17 items and computes a sample mean of 224 and a sample standard deviation of 32. She decides to construct a 98% confidence interval to estimate the mean. Assuming that the population is normally distributed, the resulting confidence interval is:

219.138 to 228.862

204.077 to 243.923

203.953 to 244.047

207.546 to 240.454

Question 10: The weights of aluminum castings produced by a process are normally distributed. A random sample of 5 castings is selected; the sample mean weight is 2.21 pounds; and the sample standard deviation is 0.12 pound. The 99% confidence interval for the population mean casting weight is:

2.009 to 2.411

2.100 to 2.320

1.825 to 2.595

1.963 to 2.457

Question 11: A researcher wants to estimate the proportion of the population which possess a given characteristic. A random sample of size 600 is taken resulting in 276 items which possess the characteristic. The point estimate for this population proportion is _______.

0.54

0.46

0.35

0.67

Question 12: A researcher wants to estimate the proportion of the population that possesses a given characteristic using a 90% confidence interval. A random sample of size 600 is taken resulting in 330 items that possess the characteristic. The error of the estimation of the confidence interval is:

0.0398

0.0068

0.0334

0.55

Question 13: To estimate the proportion of a population that possesses a given characteristic, a random sample of 1700 people are interviewed from the population. Seven hundred and fourteen of the people sampled possess the characteristic. Using this information, the researcher computes an 88% confidence interval to estimate the proportion of the population who possess the given characteristic. The resulting confidence interval is:

.401 to .439

.409 to .431

.392 to .448

.389 to .451

Question 14: From a sample of 42 items, a company wants to estimate the proportion of the population that is defective. Using the results of the sample given below, construct a 96% confidence interval to estimate that proportion. In the data below, a "y" denotes a defect.

.685 to .934

.049 to .332

.066 to .314

.072 to .309

Question 15: If a researcher is calculating a confidence interval and increases the confidence then the width of the confidence interval will do what, all other things being constant?

Remain the same

Increase

Decrease

None of the above

Question 16: The relationship of the sample variance to the population variance is captured by which distribution?

z distribution

normal distribution

t distribution

chi-square distribution

Question 17: A researcher wants to estimate the population variance. He is certain that the population is normally distributed. In an effort to construct a confidence interval, he randomly selects eight members of the population. The data are shown below. What is the point estimate of the population variance?

7.44

2.92

8.5

2.727

Question 18: A financial officer wants to estimate the population variance of daily deposits at the bank. The officer randomly records 14 deposits. The sample mean deposit is $235 with a sample standard deviation of $42. In estimating the population variance from these data, what is the point estimate?

$42

$235

$31.31

$1764

Question 19: A researcher wants to construct a 99% confidence interval to estimate a population variance using a random sample of 13 observations. The chi-square values for this confidence interval are:

4.40378, 23.3367

3.57055, 26.2170

3.56504, 29.8193

3.07379, 28.2997

Question 20: A fund manager manages a portfolio of 250 common stocks. The manager relies on various statistics, such as variance, to assess the overall risk of stocks in an economic sector. The manager's staff reported that for a sample 12 utility stocks the mean annualized return was 14% and that the variance was 3%. The 95% confidence interval for the population variance of annualized returns is

1.41 to 7.49

1.68 to 7.21

1.51 to 8.65

4.52 to 25.95

Question 21: In determining the sample size necessary to estimate a population mean, the error of estimation, E, is equal to

the distance between the sample mean and the population mean

the distance between the sample mean and the variance

the z score

the sample size

Question 22: In determining the sample size necessary to estimate p, if there is no good approximation for the value of p available, then what value should be used as an estimate of p in the formula?

0.10

0.40

0.50

1.96

Question 23: A researcher wants to estimate the average diameter of the population of a three foot long pipe. The researcher wants to be within .1" of the actual average and be 90% confident. The population variance of diameters for this type pipe is .25. How large of a sample size should be taken?

68

17

97

24

Question 24: A business researcher wants to estimate the proportion of all workers who feel stressed out with their job. The researcher is certain that the proportion is no more .22. She wants to be 99% confident of the results and be within .04 of the true population proportion. She needs to sample at least:

111

412

712

1036

Question 25: A researcher with a large national chain of fast food restaurants wants to estimate the proportion of customers who order French fries with their hamburger. The researcher is uncertain about what the proportion may actually be, wants to be 95% confident about the results, and wants to be within .03 of the actually figure. The researcher needs to sample at least:

33

1068

1842

2134

#### Solution Preview

Question 1:When a statistic calculated from sample data is used to estimate a population parameter, it is called :

an interval estimate

a point estimate

a statistical parameter

a good guess

Answer: a point estimate

Question 2: A large appliance company sends out technicians to unpack, assemble, and connect every gas dryer that is sold. In developing a pricing strategy, it is important to the company to have a "handle" on how long this process takes. The company hires a research assistant to follow technicians on 45 randomly selected jobs. The research assistant records how much time it takes the technicians to unpack, assemble, and connect each gas dryer. The resulting sample mean is 34.3 minutes. If research assistant concludes based on the sample mean that the average time for all such jobs is 34.3 minutes she is using a:

a range estimate

a statistical parameter

an interval estimate

a point estimate

Answer: a point estimate

Question 3:The Z value associated with a two sided 90% confidence interval is:

1.28

1.645

1.96

2.575

Answer:1.645

Confidence level=90%

or Significance level alpha (a) =0.1

No of tails=2

Z value corresponding to 0.1 significance level and 2 tails=1.6449

Question 4:Suppose a random sample of 36 is selected from a population with a standard deviation of 12. If the sample mean is 98, the 99% confidence interval to estimate the population mean is:

94.08 to 101.92

97.35 to 98.65

92.85 to 103.15

93.34 to 102.66

Answer:92.85 to 103.15

99%Confidence limits

Mean=M =98

Standard deviation =s=12

sample size=n=36

sx=standard error of mean=s/square root of n=2= ( 12 /square root of 36)

Confidence level=99%

Therefore Significance level=alpha (a) =1%=100% -99%

No of tails=2

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

Since sample size=36>=30

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

use normal distribution

Z at the 0.01 level of significance 2 tailed test =2.5758

Upper confidence limit=M +z*sx=103.15=98+2.5758*2

Lower confidence limit=M -z*sx=92.85=98-2.5758*2

99%Confidence limit:

Upper limit=103.15

Lower limit=92.85

Question 5:What is the average length of a wireless phone call? Suppose researchers in the telecommunications industry want to estimate the average number of minutes of a wireless call. To do so, they randomly select one such call from 85 wireless phone bills around the country. The resulting sample mean is 2.54 minutes with a standard deviation of 1.20 minutes. From these data, the 86% confidence interval to estimate the average length of a wireless phone call for all users is:

2.285 to 2.795

2.326 to 2.754

2.347 to 2.733

2.519 to 2.561

Answer:2.347 to 2.733

86%Confidence limits

Mean=M =2.54

Standard deviation =s=1.2

sample size=n=85

sx=standard error of mean=s/square root of n=0.1302= ( 1.2 /square root of 85)

Confidence level=86%

Therefore Significance level=alpha (a) =14%=100% -86%

No of tails=2

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

Since sample size=85>=30

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

use normal distribution

Z at the 0.14 level of significance 2 tailed test =1.4758

Upper confidence limit=M +z*sx=2.732=2.54+1.4758*0.1302

Lower confidence limit=M -z*sx=2.348=2.54-1.4758*0.1302

86%Confidence limit:

Upper limit=2.732

Lower limit=2.348

Question 6:The t statistic was developed by:

Charles Student

William Gosset

Abraham deMoivre

Karl Gauss

Answer:William Gosset

Question 7:In order to find values in the t distribution table, you must convert the sample size or sizes to:

population sizes

z values

student values

degrees of freedom

Answer:degrees of freedom

Question 8:A researcher is taking a random sample of 18 items in an effort to estimate the population mean. He wants to be 95% confident of his results. The table t value that he should use is:

2.11

2.101

1.74

1.734

Answer:2.11

Confidence level=95%

Or ...

#### Solution Summary

Answers multiple choice questions in statistics