Question

The cable company takes an simple random sample of 250 of the approximately 15,000 households who subscribe to them. They found that 75% of the sampled households watch sports on television at least once a month. The company is considering taking more samples like this. Suppose that it is actually 70% of their total subscribed households who watch those sports. Let p​ represent the proportion of a sample of 250 households that watch sports on television at least once a month. What is the standard deviation of the sampling distribution of p?

Answer 1

Answer:

The standard deviation of the sampling distribution of p is 0.03.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$  

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

It is provided that, a cable company takes an simple random sample of 250 of the approximately 15,000 households who subscribe to them to determine the proportion of households that watch sports on television at least once a month.

The actual proportion is 70%.

As the sample selected is quite large, i.e. n = 250 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportions.

Compute the standard deviation of this sampling distribution as follows:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

$=\sqrt{\frac{0.70(1-0.70)}{250}}\\\\=0.0289828\\\\\approx 0.03$

Thus, the standard deviation of the sampling distribution of p is 0.03.