Question

A national television network took an exit poll of 1350 voters after each had cast a...

A national television network took an exit poll of 1350 voters after each had cast a vote in a state gubernatorial election. Of? them, 610 said they voted for the Republican candidate and 740 said they voted for the Democratic candidate. Treating the sample as a random sample from the population of all? voters, a 95?% confidence interval for the proportion of all voters voting for the Democratic candidate was (0.522,0.575). Suppose the same proportions resulted from nequals=135 ?(instead of1350?), with counts of 61 and 74?, and that there are only two candidates. Complete parts a and b below.

a. Does a 95?% confidence interval using the smaller sample size allow you to predict the? winner? Explain.

The 95?% confidence interval for the proportion of all voters voting for the Democratic candidate is ( , ). Now a 95?% confidence interval (does or does not?) allow you to predict the? winner,

since this interval (includes, does not include?) (1, 0.5, 0?)

(Round to three decimal places as? needed.)

b. Explain why the same proportions but with smaller samples provide less information.? (Hint: What effect does n have on the standard? error?)

A smaller sample results in a (greater, lesser?) standard? error, which results in a (greater, lesser?) (point estimate, margin of error?) for the same proportions and confidence? level, meaning less information is provided.

Homework Answers

Answer #1

a. p = 74/135 = 0.548

95% confidence interval using the smaller sample size is given as :

Lower limit = p - z0.025 (p*(1-p) / n)0.5 = 0.548 - 1.96 *(0.548*(1 - 0.548) / 135)0.5 = 0.464

Upper limit =  p + z0.025 (p*(1-p) / n)0.5 = 0.548 + 1.96 *(0.548*(1 - 0.548) / 135)0.5 = 0.632

The 95% confidence interval for the proportion of all voters voting for the Democratic candidate is ( 0.464 , 0.632 )

Now a 95% confidence interval does not allow you to predict the winner, since this interval includes 0.5

b.

A smaller sample results in a greater standard error, which results in a greater margin of error for the same proportions and confidence level, meaning less information is provided.

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