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Chapter 7: Q2 (page 351)

ComputersIn order to better plan for student resources, the chairperson of the mathematics department at Broward College wants to estimate the percentage of students who own a computer. If we want to estimate that percentage based on survey results, how many students must we survey in order to be 90% confident that we are within four percentage points of the population percentage? Assume that the number of students is very large.

Short Answer

Expert verified

The number of students that must be surveyed to be 90% confident is 423.

Step by step solution

01

Given information

The level of confidence is 90%.

The margin of error is E=0.04.

02

Compute the sample size

The level of confidence is 90%, which implies that the level of significance is 0.10.

From the Z table, the critical value at 0.10 level of significance is 1.645.

Since no prior information on the sample proportion p^is given, the number of students that must be surveyed to be 90% confident is computed as follows:

role="math" localid="1648195763160" n=(z伪2)20.25E2=1.6452脳0.250.042=422.8鈮�423

Therefore, the number of students that must be surveyed to be 90% confident is 423.

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Most popular questions from this chapter

In Exercises 1鈥�3, refer to the accompanying screen display that results from the Verizon airport data speeds (Mbps) from Data Set 32 鈥淎irport Data Speeds鈥� in Appendix B. The confidence level of 95% was used.

Interpreting a Confidence Interval The results in the screen display are based on a 95%confidence level. Write a statement that correctly interprets the confidence interval.

Determining Sample Size. In Exercises 19鈥�22, assume that each sample is a simple random sample obtained from a normally distributed population. Use Table 7-2 on page 338 to find the indicated sample size. Quarters When setting specifications of quarters to be accepted in a vending machine, you must estimate the standard deviation of the population of quarters in use. Find the minimum sample size needed to be 99% confident that the sample standard deviation is within 10% of the population standard deviation.

In Exercises 1鈥�3, refer to the accompanying screen display that results from the Verizon airport data speeds (Mbps) from Data Set 32 鈥淎irport Data Speeds鈥� in Appendix B. The confidence level of 95% was used

Airport Data Speeds Refer to the accompanying screen display.

a. Express the con铿乨ence interval in the format that uses the 鈥渓ess than鈥� symbol. Given that the original listed data use one decimal place, round the con铿乨ence interval limits accordingly.

b. Identify the best point estimate of and the margin of error.

c. In constructing the con铿乨ence interval estimate of , why is it not necessary to con铿乺m that the sample data appear to be from a population with a normal distribution?

Chickenpox : You plan to conduct a survey to estimate the percentage of adults who have had chickenpox. Find the number of people who must be surveyed if you want to be 90% confident that the sample percentage is within two percentage points of the true percentage for the population of all adults.

a. Assume that nothing is known about the prevalence of chickenpox.

b. Assume that about 95% of adults have had chickenpox.

c. Does the added knowledge in part (b) have much of an effect on the sample size?

Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size N, and the sample size is more than 5% of the population size ,better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by N-nN-1For the sample of 100 weights of M&M candies in Data Set 27 鈥淢&M Weights鈥� in Appendix B, we get x炉=0.8656鈥�gands=0.0518鈥�g First construct a 95% confidence interval estimate of , assuming that the population is large; then construct a 95% confidence interval estimate of the mean weight of M&Ms in the full bag from which the sample was taken. The full bag has 465 M&Ms. Compare the results.
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