A market researcher would like to know how much time the average college student spends watching sports. Suppose he does a preliminary study based upon a sample of 25 college students. The mean amount of time spent watching sports, by those in the sample, is 90 minutes per day with a standard deviation of 15 minutes per day.
(a) Which of the following numbers gives the value of the point estimate?
a. 15.
b. 25.
c. 90.
d. 0.
e. 95.
(b) Using the information given above, calculate the margin of error (in minutes) of the point estimate. Assume we would like to estimate the population mean with 95% confidence.
(c) Suppose that the researcher is not content with the current margin of error. He would like the margin of error of the point estimate to be no more than 4 minutes with 95% confidence. The researcher is unwilling to trust that the sample standard deviation from his previous study accurately estimates the true population standard deviation. He does, however, know that college students tend to watch sports between 0 and 120 minutes per day. What is the minimum sample size needed to obtain this amount of precision?

(b) Using the information given above, calculate the margin of error (in minutes) of the point estimate. Assume we would like to estimate the population mean with 95% confidence.

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.0639.

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

Here, we have [tex]s = 15, n = 25[/tex]. So

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

[tex]M = 2.0639\frac{15}{\sqrt{25}}[/tex]

[tex]M = 6.19[/tex]

The margin of error of the point estimate is of 6.19 minutes.

Question c:

College students tend to watch sports between 0 and 120 minutes per day, which means that the standard deviation for population can be estimated as the standard deviation of an uniform distribution with [tex]a = 0, b = 120[/tex]. This standard deviation is:

[tex]\sigma = \sqrt{\frac{(b-a)^2}{12}}[/tex]

Then

[tex]\sigma = \sqrt{\frac{(120-0)^2}{12}}[/tex]

[tex]\sigma = 34.64[/tex]

Minimum sample size:

Now, since we have an estimate for the standard deviation of the population, we use the z-distribution.

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

The minimum sample size is n for which M = 4. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]4 = 1.96\frac{34.64}{\sqrt{n}}[/tex]

[tex]4\sqrt{n} = 1.96*34.64[/tex]

[tex]\sqrt{n} = \frac{1.96*34.64}{4}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*34.64}{4})^2[/tex]

[tex]n = 288.1[/tex]

Rounding up:

The minimum sample size needed to obtain this amount of precision is 289.