A newspaper reporter asked an srs of 100 residents in a large city for their opinion abut the mayo's job performance. using the results from the sample, the c% confidence interval for the proportion of all residents in the city who approve of the majoyr's job performance is 0.565 and 0.695. what is the value of c?

82.30%

Explanation:

Given

n = Sample Size = 100

Confidence Interval Proportion = (0.565,0.695)

The sample proportion lies at exactly the middle of the confidence interval and is calculated by:

p = (0.565 + 0.695)/2

p = 1.26/2

p = 0.63

Next, we'll solve for the boundaries of the confidence interval.

This is given by:

p ± zα/2 * √p(1-p)/n

p ± zα/2 * √p(1-p)/n = 0.695 where p = 0.63

So, we have

0.63 + zα/2 * √(0.63(1 - 0.63)/100) = 0.695 Subtract 0.63 from both sides

zα/2 * √((0.63 * 0.37)/100) = 0.695 - 0.63

zα/2 * √((0.63 * 0.37)/100) = 0.065

zα/2 * 0.048 = 0.065

zα/2 = 0.065/0.048

zα/2 = 1.346301159669266

zα/2 = 1.35 ; Approximated

The confidence interval is the probability that the sample proportion is between -zα/2 and zα/2 (-1.35 and 1.35).

This can be solved using normal probability table. Such that

Confidence Interval = P(-1.35 < Z < 1.35)

= P(Z<1.35) - P(Z<-1.35)

= 0.9115 - 0.0885

= 0.8230

= 82.30%

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