Question

Suppose that in 2017, 52% of U.S. households had only wireless telephone service (no landlines). Suppose that in 2021, 56% of a random sample of 100 U.S. households had only wireless telephone service.

1. Using the proportion from 2017, calculate the standard error. (Remember, in the modules this test covers, we usually use the earlier year's proportion as an estimate of this year's proportion when we are calculating the standard error.)

2. Does the data allow the use of a normal model for our calculations? Explain your reasoning.

3. In this part of the test, we'll go ahead and use a normal model, regardless of whether you got "yes" or "no" as your answer to part (b). Using a normal model, calculate the approximate 95% confidence interval for the percentage of all U.S. households in 2021 that have only wireless telephone service. Explain your reasoning.

Answer 1

Using proportion concepts, it is found that:

1. The standard error is of 0.05.

2. Since both $np \geq 10$ and $n(1 - p) \geq 10$, a normal model can be used.

3. The 95% confidence interval for the percentage of all U.S. households in 2021 that have only wireless telephone service is (46.27%, 65.73%).

Item 1:

The standard error for a proportion p in a sample of size n is given by:

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

For 2017, we have that:

• 52% of the households, hence $p = 0.52$
• Sample of 100 households, hence $n = 100$

Then:

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

$s = \sqrt{\frac{0.52(0.48)}{100}}$

$s = 0.05$

The standard error is of 0.05.

Item 2:

The normal model can be used if: $np \geq 10$ and $n(1 - p) \geq 10$

$np = 100(0.52) = 52 \geq 10$

$n(1 - p) = 100(0.48) = 48 \geq 10$

Since both $np \geq 10$ and $n(1 - p) \geq 10$, a normal model can be used.

Item 3:

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

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

Estimate of 2021, which was of 56%, hence $\pi = 0.56$.

95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 - 1.96\sqrt{\frac{0.56(0.44)}{100}} = 0.4627$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 + 1.96\sqrt{\frac{0.56(0.44)}{100}} = 0.6573$

As a percentage:

0.4627 x 100% = 46.27%

0.6573 x 100% = 65.73%

The 95% confidence interval for the percentage of all U.S. households in 2021 that have only wireless telephone service is (46.27%, 65.73%).

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