# A survey of 76 commercial airline flights of under 2 hours resulted

Question 3

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.37 minutes. The population standard deviation was 12 minutes. Construct a 95% confidence interval for the average time that a commercial flight of under 2 hours is late. What is the point estimate? What does the interval tell about whether the average flight is late?

*Round your answers to 2 decimal places, the tolerance is +/-0.01.

* ≤ μ ≤  *
The point estimate is  .

The interval is  . Since zero is in the interval, there is a possibility that, on average, the flights are  .

Question 5

What is the average length of a company’s policy book? Suppose policy books are sampled from 45 medium-sized companies. The average number of pages in the sample books is 213, and the population standard deviation of 48. Use this information to construct a 98% confidence interval to estimate the mean number of pages for the population of medium-sized company policy books.

Round your answers to 2 decimal places, the tolerance is +/-0.01.

≤ µ ≤

Question 7

A random sample of size 20 is taken, resulting in a sample mean of 15.40 and a sample standard deviation of 2.56. Assume x is normally distributed and use this information and α = .05 to test the following hypotheses.

Round your answer to 2 decimal places, the tolerance is +/-0.01.

Question 8

Alan Bissell, market analyst for City Sound Mart, is analyzing the relation between heavy metal CD sales and the size of the teenage population. He gathers data from six sales districts. Alan’s dependent variable is annual heavy metal CD sales (in \$1,000,000’s), and his independent variable is teenage population (in 1,000’s). Regression analysis of the data yielded the following tables.

 Coefficients Standard Error t Statistic p-value Intercept -0.14156 0.292143 -0.48455 0.653331 x 0.105195 0.013231 7.950352 0.001356 Source df SS MS F Se = 0.237 Regression 1 3.550325 3.550325 63.20809 r2 = 0.940483 Residual 4 0.224675 0.056169 Total 5 3.775

Alan’s regression model can be written as: __________.

 y = -0.14156 + 0.105195x

 y = -0.48455 + 7.950352x

 y = 0.105195 – 0.14156x

 y = 0.105195 + 0.14156x

 y = 7.950352 – 0.48455x

A random sample of 84 items is taken, producing a sample mean of 43. The population standard deviation is 5.89. Construct a 90% confidence interval to estimate the population mean.

Round your answers to 2 decimal places, the tolerance is +/-0.01.

μ

Question 15

According to the U.S. Bureau of Labor Statistics, the average weekly earnings of a production worker in 1997 were \$424.20. Suppose a labor researcher wants to test to determine whether this figure is still accurate today. The researcher randomly selects 55 production workers from across the United States and obtains a representative earnings statement for one week from each. The resulting sample average is \$430.75. Assuming a population standard deviation of \$33.90, and a 10% level of significance, determine whether the mean weekly earnings of a production worker have changed.

Round your answer to 2 decimal places, the tolerance is +/-0.01.

The value of the test statistic is   and we  .

Question 23

According to HowtoAdvice.com, the average price charged to a customer to have a 12 by 18 wall-to-wall carpet shampoo cleaned is about \$50. Suppose that a start-up carpet-cleaning company believes that in the region in which they operate, the average price for this service is higher. To test this hypothesis, the carpet-cleaning company randomly contacts 23 customers who have recently had a 12′ by 18 wall-to-wall carpet shampoo cleaned and asked the customers how much they were charged for the job. Suppose the resulting data are given below and that the population standard deviation price is \$3.49. Use a 10% level of significance to test their hypothesis. Assume that such prices are normally distributed in the population. What is the observed value? What is the p-value? What is the decision?

 \$52 52 54 50 50 51 49 49 54 51 51 48 56 52 52 53 56 52 52 56 57 49 53

*Round your answer to 2 decimal places, the tolerance is +/-0.01.
**Round your answer to 4 decimal places, the tolerance is +/-0.0001.

The value of the test statistic is  * and we  .

The p-value is  **. Since this is   than α = 0.10, the decision using the p-value is to  .

### Question 33

According to a survey by Runzheimer International, the average cost of a fast-food meal (quarter-pound cheeseburger, large fries, medium soft drink, excluding taxes) in Seattle is \$4.82. Suppose this figure was based on a sample of 27 different establishments and the standard deviation was \$0.37. Construct a 95% confidence interval for the population mean cost for all fast-food meals in Seattle. Assume the costs of a fast-food meal in Seattle are normally distributed. Using the interval as a guide, is it likely that the population mean is really \$4.50? Why or why not?

Round the answers to 4 decimal places.

μ

Since 4.50   in the interval, we are 95% confident that μ   4.50.

### Question 35

A national beauty salon chain wants to estimate the number of times per year a woman has her hair done at a beauty salon if she uses one at least once a year. The chain’s researcher estimates that, of those women who use a beauty salon at least once a year, the standard deviation of number of times of usage is approximately 6. The national chain wants the estimate to be within one time of the actual mean value. How large a sample should the researcher take to obtain a 98% confidence level?

Round your answer up to the nearest integer.

Sample  .

The tolerance is +/- 1.

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