In a poll, 597 of 1002 adults surveyed said they wanted to lose weight. a. What

Question

In a poll, 597 of 1002 adults surveyed said they wanted to lose weight. a. What percentage of the sample wanted to lose weight? b Find a 95% confidence interval for the proportion of people in the population who wanted to lose weight C. Would a 99% confidence interval be wider or narrower? a. What percentage of the sample wanted to lose weight? [1% (Round to one decimal place as needed.) b Find a 95% confidence interval for the proportion of people in the population who wanted to lose weight Round to three decimal places as needed) c Would a 99% confidence interval be wider or narrower? O A. A 99% confidence interval would be narrower because the critical value z * s greater O B A 99% confidence interval would be wider because the critical value z* is greater C. A 99% confidence interval would be narrower because the critical value z· is smaller D. A 99% confidence interval would be wider because the critical value z is smaller

Explanation / Answer

Ans:

a)Sample percentage=100*597/1002=59.6%

b)z value for 95% is 1.96

95% confidence interval for p

=0.596+/-1.96*sqrt(0.596*(1-0.596)/1002)

=0.596+/-0.030

=(0.566,0.626)

c)z value for 99% CI is 2.58,so z value is greater,hence 99% CI will be wider.

Option B is correct.

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