Chapter 7.1, Problem 9E

(a)

To determine

Whether the distribution is the sampling distribution of $\widehat{p}$ or not.

Answer to Problem 9E

The distribution is not the sampling distribution of $\widehat{p}$ .

Explanation of Solution

Given information:

Percent of students who did their assigned homework last week = 60%

Number of students for SRS = 100

The dotplot:

  

Here, the sampling distribution should have the results of all possible samples whose size is 100.

But, it is seen that the dotplot contains the results of 250 simple random samples whose size is 100.

Hence, the distribution is not the sampling distribution of $\widehat{p}$ .

(b)

To determine

The description of the distribution. Also, find whether there are any obvious outliers or not.

Answer to Problem 9E

The histogram has 4 outliers.

Explanation of Solution

Given information:

Percent of students who did their assigned homework last week = 60%

Number of students for SRS = 100

The dotplot:

  

The highest peak is roughly in the middle of the histogram; so the shape of the histogram is roughly unimodal and symmetric. Also, the highest peak of the histogram is at about 0.60. This means, the distribution is centered at 0.60. It is also seen that the values of data vary from 0.47 to 0.74 in the histogram.

Observe that there are 2 outliers on each side of the dotplot. This means the histogram has total 4 outliers and they are 0.47, 0.48, 0.73 and 0.74.

Hence, the histogram has 4 outliers.

(c)

To determine

The conclusion about the newspaper article’s claim.

Answer to Problem 9E

The newspaper article’s claim seems to be false.

Explanation of Solution

Given information:

Number of students who did all their homework last week = 45

Percent of students who did their assigned homework last week = 60%

Number of students for SRS = 100

The dotplot:

  

Consider that the number of students who did their homework last week be x and the number of students in the SRS be n.

Write:

  $x=45$

  $n=100$

Also, consider that the percent of students who did their homework last week be p.

So, $p=60\%=0.60$

Find the sample proportion by dividing the number of successes by the sample size that is dividing $x$ by $n$ .

  $\widehat{p}=\frac{x}{n}$

Substitute 45 for x and 100 for n in the above expression and simplify.

  $\begin{array}{c}\widehat{p}=\frac{45}{100}\\ =0.45\end{array}$

In the given dotplot, it is seen that there are no dots above 0.45. This means that if the population proportion is 0.60 then a sample proportion of 0.45 cannot be obtained.

Hence, the newspaper article’s claim seems to be false.