Question

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.

Answer 1

`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).

Answer 2

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).