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A marketing firm wishes to know what proportion of viewers of Impractical Jokers feels that the current season is at least as good as, or better, than previous seasons. A randomly selected group of 200 was polled. 58 responded that they felt that quality standards have been maintained. Please calculate a 90% confidence interval for the true population proportion that feels that the current season is as good as, or better, than previous seasons.

User ZeMoon
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2 Answers

5 votes

Answer:


0.29 - 1.64\sqrt{(0.29(1-0.29))/(200)}=0.237


0.29 + 1.64\sqrt{(0.29(1-0.29))/(200)}=0.343

And the confidence interval for this case would be (0.237; 0.343).

Explanation:

We can begin find the proportion estimated of responded that they felt that quality standards have been maintained with the following formula:


\hat p = (X)/(n)

And replacing we got:


\hat p =(58)/(200)= 0.29

The confidence interval is given by 90%, and the significance level would be
\alpha=1-0.90=0.1 and
\alpha/2 =0.05. And the critical value would be given by:


z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64

The confidence interval for the true proportion is given by the following formula:


\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}

Replacing the values we got:


0.29 - 1.64\sqrt{(0.29(1-0.29))/(200)}=0.237


0.29 + 1.64\sqrt{(0.29(1-0.29))/(200)}=0.343

And the confidence interval for this case would be (0.237; 0.343).

User Nrob
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6.6k points
5 votes

Answer:

The 90% confidence interval for the true population proportion that feels that the current season is as good as, or better, than previous seasons is (0.2372, 0.3428).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 200, \pi = (58)/(200) = 0.29

90% confidence level

So
\alpha = 0.1, z is the value of Z that has a pvalue of
1 - (0.1)/(2) = 0.95, so
Z = 1.645.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.29 - 1.645\sqrt{(0.29*0.71)/(200)} = 0.2372

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.29 + 1.645\sqrt{(0.29*0.71)/(200)} = 0.3428

The 90% confidence interval for the true population proportion that feels that the current season is as good as, or better, than previous seasons is (0.2372, 0.3428).

User Kadepeay
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6.7k points