Chapter 19, Problem 32CE

Christina Sanders is a member of the women’s basketball team at Windy City College. Last season, she made 55% of her free throw attempts. In an effort to improve this statistic, she attended a summer camp devoted to shooting techniques. The next 20 days she shot 100 free throws each day. She carefully recorded the number of free throws that she made each day. The results are reported below.

To interpret, the first day she made 55 out of 100, or 55%. The second day she made 61 shots, the third day, 52 shots, the tenth day 58, the eleventh day 57 shots. The last day she made 67 out of 100, or 67%.

1. a. Develop a control chart for the proportion of shots made. Over the 20 days of practice, what percent of attempts did she make? What are the upper and lower control limits for the proportion of shots made?
2. b. Is there any trend in her proportion made? Does she seem to be improving, staying the same, or getting worse?
3. c. Find the percent of attempts made for the last 5 days of practice. Use the hypothesis testing procedure, formula (15-1), to determine if there is an improvement from 55%.

To determine

Create a control chart for the proportion of shots made.

Find the overall percentage of shots made in the 20 days of practice.

Find the upper and lower control limits of the proportion of shots made.

Answer to Problem 32CE

The control chart for the proportion of shots made is as follows:

The overall percentage of shots made in the 20 days of practice is 61.85%.

The upper control limit of the proportion of shots made is 0.7642.

The lower control limit of the proportion of shots made is 0.4728.

Explanation of Solution

Calculation:

Denote Y as the random variable representing the number of shots made on a certain practice day, out of the 100 free throws. Thus, for each day, the number of subgroups for the construction of the control chart is 100.

Step-by-step procedure to construct the control chart for the proportion using MINITAB software is as follows:

• Choose Stat > Control Charts > Attributes Charts > P.
• In Variables, enter columns Y.
• In Subgroup size, enter 100.
• Clock OK.

The control chart for the proportion of shots made is obtained using MINITAB software.

In the obtained control chart, the central line represents the overall proportion of shots made in the 20 days of practice, which is, 0.6185.

Thus, the overall percentage of shots made in the 20 days of practice is 61.85% $\left(=100\times 0.6185\right)$.

The quantities UCL and LCL in the control chart represent the upper and lower control limits of the proportions.

Thus, the upper control limit of the proportion of shots made is 0.7642; the lower control limit of the proportion of shots made is 0.4728.

To determine

Describe the trend in the proportion of shots made, and note whether the player is showing any improvement in her shots, staying the same, or worsening over the practice period.

Answer to Problem 32CE

The player appears to be showing improvement in her shots over the practice period.

Explanation of Solution

A careful observation of the control chart for the proportion of shots made reveals that for the first 11 days, the proportions fluctuate over a wide range of values, sometimes being higher than the overall proportion of 0.6185, and sometimes lower than 0.6185. However, starting from the 12^th day and till the 20^th day, the proportion never falls below the central line of 0.6185, even though there are some fluctuations.

Thus, the player appears to be showing improvement in her shots over the practice period.

To determine

Calculate the percent of attempts made in the last 5 days of practice.

Perform a hypothesis test to decide whether there is an improvement from 55%.

Answer to Problem 32CE

The percent of attempts made in the last 5 days of practice is 65.4%.

There is a significant improvement in the shots made by the player from 55%.

Explanation of Solution

Denote $\pi$ as the true proportion of shots made in the last 5 days of practice. The null and alternate hypotheses for testing whether there is an improvement from 55% in the last 5 days of practice are given below:

Null hypothesis:

$H_0:\pi \le 0.55$, that is, there is no improvement from 55% in the last 5 days of practice.

Alternate hypothesis:

$H_1:\pi >0.55$, that is, there is an improvement from 55% in the last 5 days of practice.

This is a right-tailed test, due to the greater than-type alternate hypothesis.

The z-test for a single proportion is to be performed in this case. Assume that the level of significance is 0.05. The critical value is such that, $P\left(Z\ge z\right)=0.05$. Consider the following calculation:

$\begin{array}{c}P\left(Z\ge z\right)=0.05\\ P\left(Z<z\right)=1-P\left(Z\ge z\right)\\ =1-0.05\\ =\mathrm{0.95.}\\ P\left(-\infty <Z<0\right)+P\left(0\le Z<z\right)=0.95\\ P\left(0\le Z<z\right)=0.95-0.5\left[\because P\left(-\infty <Z<0\right)=0.5\right]\\ =\mathrm{0.45.}\end{array}$

Use Table B.3: Areas under the Normal Curve to obtain the value of z, such that $P\left(0\le Z<z\right)=0.45$, as follows:

• Locate the value 0.4500 from the body of the table. The exact value 0.4500 is not present in the table. Hence, locate the nearest values to it, as 0.4495 and 0.4505.
• Identify the value of z corresponding to both 0.4495 and 0.4505 as 1.6.
• Identify the column 0.04 for 0.4495, and the column 0.05 for 0.4505, to represent the second decimal place.
• The z-values 1.64 and 1.65 are the nearest possible values, such that, $P\left(0\le Z<1.64\right)=0.4495$ and $P\left(0\le Z<1.65\right)=0.4505$.

Since the required probability is, $0.45=\frac{0.4495+0.4505}{2}$, it can be said that $P\left(0\le Z<1.645\right)=0.45$ so that $1.645=\frac{1.64+1.65}{2}$.

Decision rule:

Reject $H_0$ if $z\ge 1.645$. Otherwise, fail to reject $H_0$.

The total number of free throws attempted in the last 5 days of practice is $500\left(=5\times 100\right)$.

The number of shots made in the last 5 days of practice is $327\left(=65+63+68+64+67\right)$.

Thus, the proportion of shots made in the last 5 days of practice is calculated below:

$\begin{array}{c}p=\frac{327}{500}\\ =\mathrm{0.654.}\end{array}$

Using the formula (15- 1), the test statistic can be calculated as follows:

$\begin{array}{c}z=\frac{p-\pi }{\sqrt{\frac{\pi \left(1-\pi \right)}{n}}}\\ =\frac{0.654-0.55}{\sqrt{\frac{0.55\left(1-0.55\right)}{500}}}\\ =\frac{0.104}{\sqrt{\frac{0.2475}{500}}}\\ \approx \mathrm{4.674.}\end{array}$

Conclusion:

The calculated z-value is greater than the critical value.

In other words, $z\left(=4.674\right)>1.645$.

Thus, according to the decision rule, reject $H_0$.

Hence, it can be concluded that there is a significant improvement in the shots made by the player from 55%.