Why is 90 the point estimate and how to find the t-critical for the following problems?

Question

Not sure how 90 was achieved for this problem for the point estimate. Also quite lost on (b) and (c) would truly appreciate the help, thank you

QUESTION:

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? SOLUTION: 90, not sure why

(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. (Use a table or technology. Round your answer to two decimal places.)

(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 5 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? (Use a table or technology.)

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Answer 1

This is a one-sample problem. As @WaveX comments, the formula mentioned in your comment is for a two-sample problem.

You have a sample of size $n = 25$ from a normal distribution with unknown mean $\mu$ and unknown standard deviation $\sigma.$ Point estimates of $\mu$ and $\sigma$ are $\bar X = 90$ and $S = 15,$ respectively. [Both estimates are used below.]

(2) The margin of error in a 95% t confidence interval is $t^* S/\sqrt{n} = 2.062(3) = 6.19,$ where $t^*$ cuts 2.5% from the upper tail of Student's t distribution with $n-1 = 24$ degrees of freedom.

(3) This part assumes that $n$ will be large enough to use a normal approximation. The margin of error of a 95% z confidence interval is $M = z^* \sigma/\sqrt{n} \approx 1.96(15)/\sqrt{n}.$ For $M = 5,$ the equation becomes $\sqrt{n} = 1.96(15)/5 = 5.88,$ so $n = 34.5744,$ which we round up to 35 students.

I think this ought to be a good start. I leave it to you to check my values $t^*$ and $z^*$ in tables or using software. Also, maybe you need to fill in a couple of small between my equations. Finally, you should make sure you understand the logic of parts (2) and (3). Ask if you think you find mistakes or still have questions. [And thanks for the transcription.]

Addendum from Minitab: For (2):

One-Sample T 

 N   Mean  StDev  SE Mean      95% CI
25  90.00  15.00     3.00  (83.81, 96.19)

Margin of error is half the length of the CI.

Checking (3): Trying $n = 35$ with t interval (no normal approximation). Margin of error seems really close to 5.

One-Sample T 

 N   Mean  StDev  SE Mean      95% CI
35  90.00  15.00     2.54  (84.85, 95.15)

And with requested z interval:

One-Sample Z 

The assumed standard deviation = 15

 N   Mean  SE Mean      95% CI
35  90.00     2.54  (85.03, 94.97)