Question

Time required to complete a task. When a person is asked, “How much time will you require to complete this task?” cognitive theory posits that people (e.g., a business consultant) will typically underestimate the time required. Would the opposite theory hold if the question was phrased in terms of how much work could be completed in a given amount of time? This was the question of interest to researchers writing in Applied Cognitive Psychology (Vol. 25, 2011). For one study conducted by the researchers, each in a sample of 40 University of Oslo students was asked how many minutes it would take to read a 32-page report. In a second study, 42 students were asked how many pages of a lengthy report they could read in 48 minutes. (The students in either study did not actually read the report.) Numerical descriptive statistics (based on summary information published in the article) for both studies are provided in the accompanying table.

a. The researchers determined that the actual mean time it takes to read the report in\(\mu = 48\) minutes. Is there evidence to support the theory that the students, on average, overestimated the time it would take to read the report? Test using\(\alpha = 0.10\).

b. The researchers also determined that the actual mean number of pages of the report that is read within the allotted time is\(\mu = 32\)pages. Is there evidence to support the theory that the students, on average, underestimated the number of report pages that could be read? Test using\(\alpha = 0.10\)

c. The researchers noted that the distribution of both estimated time and number of pages is highly skewed (i.e., not normally distributed). Does this fact impact the inferences derived in parts a and b? Explain.

Short answer

a. It may conclude that the student will overestimate the time taken to read the report.

b.It may conclude that the student, on average, underestimated the time number of pages that could be read.

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

Step-by-step solution

Given information

The given data is as follows:

(a) Testing the hypothesis for \(\mu  = 48\)

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.

(b) Testing the hypothesis for \(\mu  = 32\)

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.

(c) Explaining the given statement

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).