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The given data is as follows:

Let\(\mu \)denote the estimated mean time to read the report. The main objective is to test whether the student will overestimate the time taken to read the report or not.

The null hypothesis for the test is given below:

\({H_0}:\mu = 48\)

That is, the time taken to read the report is equal to 48 minutes.

The alternative hypothesis for the test is given below:

\({H_a}:\mu > 48\)

That is, the time taken to read the report is more than 48 minutes.

Consider\(\overline x = 60,\sigma = 41\)and\(n = 40\)

The value of the test statistic\(\left( z \right)\)is obtained below:

\(\begin{aligned}z &= \frac{{\overline x - \mu }}{\sigma }\\ &= \frac{{60 - 48}}{{\frac{{41}}{{\sqrt {40} }}}}\\ &= 1.85\end{aligned}\)

Let, from the standard normal table,

\(\begin{aligned}p &= P\left( {z \ge 1.85} \right)\\ &= 0.5 - 0.4678\\ &= 0.0322\end{aligned}\)

Thus, the required p-value is 0.0322.

Since the p-value is less than \(\alpha \) , therefore, the null hypothesis is rejected, hence, it may conclude that the student will overestimate the time taken to read the report.

Let\(\mu \)denote the estimated mean time of the number of pages to be read. The main objective is to test whether the student, on average, underestimated the number of pages that could be read from the actual meantime or not.

The null hypothesis for the test is given below:

\({H_0}:\mu = 32\)

That is, the student, on average, takes 32 minutes to read the number of pages.

The alternative hypothesis for the test is given below:

\({H_a}:\mu < 32\)

That is, the average student underestimated the number of pages that could be read.

Consider\(\overline x = 28,\sigma = 14\)and\(n = 42\)

The value of the test statistic\(\left( z \right)\)is obtained below:

\(\begin{aligned}z &= \frac{{\overline x - \mu }}{\sigma }\\ &= \frac{{28 - 32}}{{\frac{{14}}{{\sqrt {42} }}}}\\ &= - 1.85\end{aligned}\)

Let, from the standard normal table,

\(\begin{aligned}p &= P\left( {z \le - 1.85} \right)\\ &= 0.5 - 0.4678\\ &= 0.0322\end{aligned}\)

Thus, the required p-value is 0.0322.

Since the p-value is less than,\(\alpha \) the null hypothesis is rejected; hence, it may conclude that the student, on average, underestimated the number of pages that could be read.

According to Central Limit Theorem, if n is large, then \(\overline x \) it follows the normal distribution with the mean \({\mu _{\overline x }} = \mu \) and the standard deviation \({\sigma _{\overline x }} = \frac{\sigma }{{\sqrt n }}\)

Thus,

The sampling distribution\(\overline x \)for both tests is of normal shape.

Since, for both tests, the sample size is greater than 30.

Hence, the fact that the estimated time and the number of pages are highly skewed will not affect the inferences derived in parts a) and parts b).