Question

The mayor of a city believes a large park in the city is becoming more popular and wants to expand nature programs at the park. A random sample of 30 summer days are selected and the number of park visitors is determined for each of those days. From this sample, the mean number of visitors was X(xbar) = 98.3 with a standard deviation of Sx = 11.2. Calculate a 99% confidence interval for the true mean number of visitors on summer days. Assume the conditions for inference are met.
a. 95.192 to 101.410 visitors
b. 34.123 to 162.480 visitors
c. 92.664 to 103.940 visitors
d. None of these is correct.

Answer 1

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 99% confidence interval for the true mean number of visitors on summer days is given by:

c. 92.664 to 103.940 visitors

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 30 - 1 = 29 df, is t = 2.756.

The other parameters are given as follows:

$\overline{x} = 98.3, s = 11.2, n = 30$.

Hence, the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 98.3 - 2.756\frac{11.2}{\sqrt{30}} = 92.664$

$\overline{x} + t\frac{s}{\sqrt{n}} = 98.3 + 2.756\frac{11.2}{\sqrt{30}} = 103.94$

Hence option c is correct.

More can be learned about the t-distribution at brainly.com/question/16162795