On March 9-10, 1989, the Newsweek Poll interviewed what was called a national sa

Question

On March 9-10, 1989, the Newsweek Poll interviewed what was called a national sample of 756 adults by telephone. Newsweek magazine reported that 469 respondents "approve of the way President George Bush is handling his job." Assume that the 756 respondents were selected at random from all adults living in the United States. Test the null hypothesis that the percentage of the U.S. population that approved of President Bush's handling of his job was equal to 0.50 versus the alternative that it was larger. Use the 0.05 level of significance.

Explanation / Answer

n = 756

p=469/756

q = 287/756

Sigma =sqrt( pq) = sqrt(469*287/(756*756)) = 0.48

Then, we do a single-sample t-test for comparing population means

The following results are obtained,

T-value = 6.7 using calulation (0.62-0.50)/[0.48/sqrt(756)]

p-value = 0.0003

Since,p-value is less than 0.05, we can reject the null hypothesis

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