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

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).

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 200, \pi = \frac{58}{200} = 0.29$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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).

Answer 2

Answer:

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

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

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

Step-by-step explanation:

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

$\hat p = \frac{X}{n}$

And replacing we got:

$\hat p =\frac{58}{200}= 0.29$

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:  

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

Replacing the values we got:

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

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

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