Question

A research program used a representative random sample of men and women to gauge the size of the personal network of older adults. Each adult in the sample wasasked to "please name the people you have frequent contact with and who are also important to you." The responses of 2,816 adults in this sample yielded statisticson network size, that is, the mean number of people named per person was x = 14.7, with a standard deviation of s = 10.4. Complete parts a through d.a. Give a point estimate for H.14.7 (Type an integer or a decimal.)b. Give an interval estimate for p. Use a confidence coefficient of 0.95.(Round to two decimal places as needed.)

Answer 1

Given:

Number (n) of adults sampled = 2816

mean = 14.7

standard deviation (s) = 10.4

The interval estimate using a confidence coefficient of 0.95

The interval estimate can be calculated using the formula:

$\text{Interval estimate = mean }\pm z(\frac{s}{\sqrt[]{n}})^{}$

where:

z is the z-score for the given confidence interval

From the z-score/confidence coefficient table, the z-score for a confidence coefficient of 0.95 is 1.96

The table is shown below:

When we substitute into the formula, we have:

$\begin{gathered} \text{Interval estimate = 14.7 - 1.96(}\frac{10.4}{\sqrt[]{2816}})\text{ to 14.7 + 1.96(}\frac{10.4}{\sqrt[]{2816}}) \\ =\text{ }14.7\text{ - 0.38413 to 14.7 + 0.38}413 \\ \approx\text{ 14.32 to 15.08} \end{gathered}$

(d)