Question

An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 23 homes recently sold in the Buffalo, New York, area revealed the mean number of times a home was shown was 24 and the standard deviation of the sample was 9 people.
a. What is the margin of error for a 98% confidence interval?

b.What is the 98% confidence interval for the population mean?

Answer 1

Using the t-distribution, it is found that:

a. The margin of error is of 4.7 homes.

b. The 98% confidence interval for the population mean is (19.3, 28.7).

The information given in the text is:

• Sample mean of $\overline{x} = 24$.
• Sample standard deviation of $s = 9$.
• Sample size of $n = 23$.

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The confidence interval is:

$\overline{x} \pm M$

The margin of error is:

$M = t\frac{s}{\sqrt{n}}$

Item a:

The critical value, using a t-distribution calculator, for a two-tailed 98% confidence interval, with 23 - 1 = 22 df, is t = 2.508.

Then, the margin of error is:

$M = t\frac{s}{\sqrt{n}} = 2.508\frac{9}{\sqrt{23}} = 4.7$

Item b:

The interval is:

$\overline{x} - M = 24 - 4.7 = 19.3$

$\overline{x} + M = 24 + 4.7 = 28.7$

The 98% confidence interval for the population mean is (19.3, 28.7).

A similar problem is given at brainly.com/question/15180581