1.The report of the sample survey of 1014 adults says with 95% confidence between 9% and 15% of all americans expect to spend more on gifts this year than last year

what does the phase 95% confidence mean?

a. there is a 95% change that the percent who expect to spend more is between 9% and 15%

b. 9% and 15% will spend 95% of what was spent last year

c. 95% will spend between 9% and 15% more than last year

d. the method used to get 9% and 15% when used over and over again produce intervals which include true population proportion 95% of the time

2.what is the critical value of t for a 90% confidence interval with df=14? (assume a 2-tailed test)

a. 1.761310

b. 1.345030

c. 2.62449

d. 2.14479

3.what is the critical value for a 95% confidence interval based on a sample size of 20? (population standard deviation is unknown) (assume a 2-tailed test)

a. 1.724718

b. 1.729133

c. 2.08596

d. 2.09302

4.a survey of 30 random shoppers found a 95% confidence interval that shoppers will spend between $24.50 and $28.30. A second survey surveyed 25 random shoppers

a. the second interval will be wider because the sample size is smaller creates uncertainty

b. the second interval will be narrower because the sample size is smaller creates less uncertainty

c. the second interval will be narrower because it is likely there are less outliers

c. the second interval will be wider because it is likely that there are more outliers

5.A survey of 531 citizens found that 303 of them savor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower level of the interval (round to 3 decimal digits)

HINT: used 68-95-99% rule

6.Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28,33,34,25,26, and 135. Which table do we use to find margin of error

a. z-table

b. t-table

c. g-table

d. f-table

7.Compute a 95% confidence interval for the population mean, based on the sample numbers 21,28,33,34,25,26, and 135. What is the critical value ? (get 4 decimal digits)

a. 2.5706

b. 3.7074

c. 2.4469

d. 2.3646

HINT: EXCEL critical value in z table = NORM.S.INV and T.INV for a T Table (Be careful, the alpha is halved in the T if it is a two tail problem)

8.Compute a 95% confidence interval for the population mean, based on the sample numbers 21,28,33,34,25,26, and 135. What is the margin of error?

a. 29.57

b. 37.69

c. 38.80

d. 41.83

9.Compute a 95% confidence interval for the population mean based on the sample of numbers 21,28,33,34,25,26, and 135. Find the lower and upper limits of the interval

a. lower 15.449 upper 67.174

b. lower 20.492 upper 40.876

c. lower 18.925 upper 50.741

d. lower 5.449 upper 80.837

10.Change the last value to 27 and re-compute the confidence interval, the numbers are now 21,28,33,34,25,26, and 27

a. lower 3.52 upper 41.929

b. lower 23.52 upper 31.909

c. lower 22.32 upper 35.861

d. lower 19.73 upper 51.994

11.What is an outlier and how does it affect the confidence interval

a. an outlier stretches the interval because it increases the standard deviation

b. an outlier compacts the interval because it increase the standard deviation

c. an outlier stretches the interval because it decrease the standard deviation

d. an outlier compacts the interval because it decrease the standard deviation

HINT: an outlier is a value that is inconsistent with the rest of the data usually defined as a value that is more than 1.5 times the interquartile range larger than Q3 or smaller than Q1.

12.A sample of 25 students enrolled in the university indicates x(bar) = $315.40 and s=$43.20. Using a significance level of 0.1, what are the boundaries of the confidence interval

a. 302.438 - 326.184

b. 300.618-330.182

c. 297.485-327-820

d. 282.271-338.475

13.A sample of 25 students enrolled in the university indicates that x(bar) = $315.40 and s=$43.20. A mean is not a confidence interval is rejected by the confidence interval and we say the evidence against the mean is significant. at the 0.10 level of significance, is there evidence against mean $300

a. no because 300 is below the lower limit of the confidence interval

b. yes because 300 is below the lower limit of the confidence interval

c. no because 300 is in the confidence interval

d. yes because 300 is in the confidence interval

14.A sample of 25 students enrolled in the university indicated that x(bar)=$315.40 and s=$75 (note the sample standard deviation is different from the first part). A mean that is ina. confidence is not rejected by the confidence interval and we say the evidence against the mean is not significant. At the 0.05 level of significance, is there evidence against the mean $300?

a. no because 300 is below the lower limit confidence interval

b. yes because 300 is below the lower limit confidence interval

c. no because 300 is in the confidence interval

d. yes because 300 is in the confidence interval

15.Explain the difference between parts A and C. In both parts we asked whether or not the evidence that the population mean was over $400.

In part A x(bar)=$315.40 s=$43.20, sample size=25

In part C x(bar)= $315.40, s=$75, sample size =25

a.the larger mean adds uncertainty in the sample, the same may have outliers

b.the smaller means adds uncertainty in the sample, the sample may have outliers

c. the larger standard deviation adds uncertainty in the sample, the same may have had outliers

d. the smaller standard deviation adds uncertainty in the sample, the sample may have had outliers.