Question

In: Statistics and Probability

Suppose that a safety group surveyed 1,300 drivers. Among those​ surveyed, 67% said that careless or...

Suppose that a safety group surveyed

1,300

drivers. Among those​ surveyed,

67%

said that careless or aggressive driving was the biggest threat on the​ road, and

37

said that cell phone usage by other drivers was the driving behavior that annoyed them the most. Based on these data and assuming that the sample was a simple random​ sample, construct and interpret a

99​%

confidence interval estimate for the true proportion in the population of all drivers who are annoyed by cell phone users.

The confidence interval estimate is ​(Round to three decimal places as needed. Use ascending​ order.)

Interpret the confidence interval estimate.

A.There is a

0.99

probability that the sample proportion of drivers who are annoyed by cell phone users is in the interval.

B.There is

99​%

confidence that the population proportion of drivers who are annoyed by cell phone users is in the interval.

C.There is a

0.99

probability that the population proportion of drivers who are annoyed by cell phone users is in the interval.

D.There is

99​%

confidence that the population proportion of drivers who are annoyed by cell phone users is equal to one of the bounds of the interval

A pharmaceutical company operates retail pharmacies in 10 eastern states.​ Recently, the​ company's internal audit department selected a random sample of

300 prescriptions issued throughout the system. The objective of the sampling was to estimate the average dollar value of all prescriptions issued by the company. The data collected were

x overbarxequals=​$14.45 and sequals=5.00 Complete parts a and b below.

a. Determine the​ 90% confidence interval estimate for the true average sales value for prescriptions issued by the company. Interpret the interval estimate.

The​ 90% confidence interval is ​(Round to the nearest cent as needed. Use ascending​ order.)

Interpret the interval. Choose the correct answer below.

A.

The company believes that the true mean prescription amount falls between these two values​ 90% of the time.

B.

There is a 0.90 probability that the true mean prescription amount is between these two values.

C.

The company believes with​ 90% confidence that the true mean prescription amount is between these two amounts.

D.

The company believes with​ 90% confidence that the sample mean prescription amount is between these two amounts.

b. One of its retail outlets recently reported that it had monthly revenue of

​$7 485 from 531

prescriptions. Are such results to be​ expected? Should that retail outlet be​ audited?Assuming the population mean is at the upper limit of the​ 90% confidence interval computed in part​ a, the upper limit of the​ 90% confidence interval for the expected total monthly revenue for

531

prescriptions would be

​$nothing.

Assuming the population mean is at the lower limit of the​ 90% confidence interval computed in part​ a, the lower limit of the​ 90% confidence interval for the expected total monthly revenue for

531

prescriptions would be

Given that this outlet reported sales of

​$7, 485 from 531

​prescriptions, there is good or no

reason to believe that this is out of line. The retail outlet

should not or should be audited.

​(Round to the nearest cent as​ needed.)

Solutions

Expert Solution

Answering the first question on cell phones :-

We need to construct the 99% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X = 37
Sample Size N = 1300

The sample proportion is computed as follows, based on the sample size N=1300 and the number of favorable cases X = 37:

The critical value for α=0.01 is . The corresponding confidence interval is computed as shown below:

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.017<p<0.04, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.017,0.04).

B.There is 99​% confidence that the population proportion of drivers who are annoyed by cell phone users is in the interval.

Please let me know in comments if anything is not clear. Will reply ASAP. Please do upvote if satisfied!!


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