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Chapter 8: Q 6. (page 507)

Losing weight A Gallup poll asked a random sample of U.S. adults, 鈥淲ould you like to lose weight?鈥� Based on this poll, the 95%confidence interval for the population

proportion who want to lose weight is (0.56,0.62).

a. Interpret the confidence interval.

b. What is the point estimate that was used to create the interval? What is the margin of error?

c. Based on this poll, Gallup claims that more than half of U.S. adults want to lose weight. Use the confidence interval to evaluate this claim.

Short Answer

Expert verified

a. There is 95%assurance that US adults who wish to lose weight lie in given interval.

b. Point estimation is 0.59and margin of error is 0.03.

c. It is a convincing evidence.

Step by step solution

01

Given Information

It is given that 95%confidence interval for proportion is(0.56,0.62)

02

Concept Used

Formula Used arep^=UCL+LCL2andME=UCL-LCL2

03

a. Interpreting Confidence Interval

It refers to 95%assurance that US adults proportion who wish to lose weight lie between0.56-0.62. Others do not wish to lose weight.

04

b. Point Estimate and Margin of Error

Point Estimation: It is given by p^=0.56+0.622=0.59

It lies halfway of confidence interval. Point estimate is average of boundaries of given interval.

Margin of Error: It is given by ME=0.62-0.562=0.03

It is half the difference of confidence interval.

05

c. Evaluating the claim

After the above observation and given data, it can be observed that population proportion who wish to lose weight is 0.50.

Hence, there is a convincing evidence that more than half of US adults who wish to lose weight.

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Most popular questions from this chapter

A quality control inspector will measure the salt content (in milligrams) in a random

sample of bags of potato chips from an hour of production. Which of the following would result in the smallest margin of error in estimating the mean salt content 渭?

a. 90%confidence; n=25

b. 90%confidence; n=50

c. 95%confidence; n=25

d. 95%confidence; n=50

e. n = 100 at any confidence level

Lying online Many teens have posted profiles on sites such as Facebook. A sample survey asked random samples of teens with online profiles if they included false information in their profiles. Of 170younger teens (ages role="math" localid="1654194610029" 12to14) polled, 117said 鈥淵es.鈥� Of 317older teens (ages 15to17) polled, 152said 鈥淵es.鈥� A 95%confidence interval for pY-pO=the true difference in the proportions of younger teens and older teens who include false information in their profile is 0.120to 0.297.

a. Interpret the confidence interval.

b. Does the confidence interval give convincing evidence of a difference in the true

proportions of younger and older teens who include false information in their profiles?

Explain your answer.

After deciding on a 95% confidence level, the researcher is deciding between a sample of size n=500 and a sample of size n=1000. Compared with using a sample size of n=500, a confidence interval based on a sample size of n=1000 will be

a. narrower and would involve a larger risk of being incorrect.

b. wider and would involve a smaller risk of being incorrect.

c. narrower and would involve a smaller risk of being incorrect.

d. wider and would involve a larger risk of being incorrect.

e. narrower and would have the same risk of being incorrect.

America鈥檚 favorite cookie Ann and Tori wanted to estimate the average weight of an Oreo cookie to determine if it was less than advertised (34 grams for 3 cookies). They selected a random sample of 36cookies and found the weight of each cookie (in grams). The mean weight was x虅=11.3921grams with a standard deviation of sX=0.0817grams. Construct and interpret a 90%confidence interval for the true mean weight of an Oreo cookie.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion

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