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Chapter 9: 54 (page 564)

Losing weight A Gallup Poll found that 59% of the people in its sample said 鈥淵es鈥� when asked, 鈥淲ould you like to lose weight?鈥� Gallup announced: 鈥淔or

results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is 魛伅3 percentage points.鈥�16 Can we use this interval to conclude that the actual proportion of U.S. adults who would say they want to lose weight differs from 0.55? Justify your answer.

Short Answer

Expert verified

No, can not be concluded as 0.55 does not lie within the confidence.

Step by step solution

01

Step 1. Calculate the confidence interval.

It is given that the proportion of people who would like to lose weight is equal to 0.59. The margin of error is equal to 卤3. The confidence interval is given as follows:

CI=0.59鈭�3,0.59+3=0.56,0.62

02

Step 2. Interpretation.

It can be observed that 0.55 does not lie within the confidence interval representing insufficient evidence to conclude that the actual proportion of US adults who will lose weight differs from 0.55.

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Most popular questions from this chapter

Refer to Exercise 2. For Yvonne's survey, 96 students in the sample said they rarely or never argue with friends. A significance test yields a P-value of 0.0291.

(a) Interpret this result in context.

(b) Do the data provide convincing evidence against the null hypothesis? Explain.

Bottles of a popular cola are supposed to contain 300milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. From experience, the distribution of the contents is approximately Normal. An inspector measures the contents of six randomly selected bottles from a single day鈥檚 production. The results are 299.4297.7301.0298.9300.2297.0Do these data provide convincing evidence that the mean amount of cola in all the bottles filled that day differs from the target value of 300ml? Carry out an appropriate test to support your answer

Stating hypotheses State the appropriate null and alternative hypotheses in each of the following cases.

(a) The average height of 18-year-old American women is 64.2inches. You wonder whether the mean height of this year's female graduates from a large local high school (over 3000students) differs from the national average. You measure an SRS of 48female graduates and find that X=63.1inches.

(b) Mr. Starnes believes that less than 75%of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of students at the school to help Mr. Starnes test his claim.

- Check conditions for carrying out a test about a population proportion or mean.

- Interpret P-values in context.

A Gallup Poll report on a national survey of 1028 teenagers revealed that 72% of teens said they seldom or never argue with their friends. Yvonne wonders whether this national result would be true in her large high school. So she surveys a

a random sample of 150 students at her school.

A government report says that the average amount of money spent per U.S. household per week on food is about \(158. A random sample of 50households in a small city is selected, and their weekly spending on food is recorded. The Minitab output below shows the results of requesting a confidence interval for the population mean M. An examination of the data reveals no outliers.

(a) Explain why the Normal condition is met in this case.

(b) Can you conclude that the mean weekly spending on food in this city differs from the national figure of\)158? Give appropriate evidence to support your answer.
See all solutions

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