Question

Our basketball player in Exercise 39 has new sneakers, which he thinks improve his game. Over his past 40 shots, he's made 32 -much better than the \(55 \%\) he usually shoots. Do you think his chances of making a shot really increased? In other words, is making at least 32 of 40 shots really unusual for him? (Do you think it's his sneakers?)

Short answer

The player's improved performance is not by chance; the new sneakers likely increased his shooting percentage.

Step-by-step solution

State the Hypotheses

We start by setting up our hypotheses. The null hypothesis \(H_0\) is that the player's shooting percentage is still 55%, while the alternative hypothesis \(H_a\) is that his shooting percentage has increased with the new sneakers.

Identify the Test Distribution

Since the number of shots made follows a binomial distribution with parameters \(n = 40\) (the number of trials) and \(p = 0.55\) (probability of success), we will use the normal approximation for the binomial distribution to evaluate the probability of making at least 32 shots.

Calculate the Mean and Standard Deviation

To use the normal approximation, calculate the mean \(\mu\) and the standard deviation \(\sigma\) of the distribution. The mean \(\mu = np = 40 \times 0.55 = 22\) and the standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{40 \times 0.55 \times 0.45} \approx 3.14\).

Compute the Z-score

Find the Z-score for making 32 shots. The Z-score is given by \(Z = \frac{X - \mu}{\sigma}\), where \(X = 32\). So, \(Z = \frac{32 - 22}{3.14} \approx 3.18\).

Determine the P-value

Using the Z-score, we find the p-value, which is the probability that the player would make 32 or more shots under the null hypothesis. Looking up \(Z = 3.18\) in a standard normal distribution table gives a very small p-value (much less than 0.05).

Make a Decision

Since the p-value is very small, we reject the null hypothesis. This suggests that it would be unusual for the player to make at least 32 shots if his shooting ability had not improved.

Key concepts

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials of a binary outcome. Each trial can result in just two possible outcomes: success or failure. The probability of success is constant across all trials. In our basketball player's case, a success is making a shot.
The distribution is defined by two parameters:

• The number of trials, denoted as \( n \). In this case, \( n = 40 \) since the player takes a total of 40 shots.
• The probability of success, denoted as \( p \). Here, \( p = 0.55 \), representing the player's usual shooting percentage.

The binomial probability of any given number of successes \( k \) out of \( n \) trials is calculated using the formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes out of \( n \) trials. Understanding a binomial distribution helps in calculating probabilities concerning the number of shots made.

Normal Approximation

Often, calculating exact probabilities using a binomial distribution can be complex, especially when the number of trials is large. This is where the normal approximation to the binomial distribution is handy. The idea is to approximate the binomial distribution with a normal distribution when certain conditions are met:

• The number of trials \( n \) is large.
• Both \( np \) and \( n(1-p) \) are at least 5.

In our exercise, \( n = 40 \) and \( p = 0.55 \), so both \( np = 22 \) and \( n(1-p) = 18 \) satisfy these conditions. Therefore, we use a normal distribution to estimate probabilities.
This normal distribution requires calculating the mean \( \mu = np = 22 \) and the standard deviation \( \sigma = \sqrt{np(1-p)} \approx 3.14 \).
The normal curve can then be used to estimate the probability of the player making at least 32 shots, simplifying the solution process.

P-value

The p-value is a statistical measure that helps determine the significance of your results in a hypothesis test. It represents the probability of observing test results as extreme as those observed, under the assumption that the null hypothesis is true.
In the context of this basketball shooting hypothesis test, we aim to find out whether making 32 or more shots, given his usual success rate is 55%, is statistically significant.

• Using our calculated Z-score of \( 3.18 \), we look up the p-value in a standard normal distribution table.
• The p-value derived from a Z-score of \( 3.18 \) is very small, far below the usual significance level of 0.05.

This small p-value allows us to conclude that the event is rare when assuming the null hypothesis is true, indicating that the player's improved shooting performance with the new sneakers is indeed statistically significant.

Z-score

The Z-score, in this context, measures how many standard deviations an element is from the mean. It's an easy way to visualize the player's shooting performance in relation to his usual shooting trend.
We calculate the Z-score using the formula:\[Z = \frac{X - \mu}{\sigma}.\]Here:

• \( X = 32 \), the number of successful shots he made.
• \( \mu = 22 \), the mean or expected number of successful shots.
• \( \sigma \approx 3.14 \), the standard deviation.

By substituting these values, we compute:\[Z = \frac{32 - 22}{3.14} \approx 3.18.\]A positive Z-score of approximately 3.18 indicates that making 32 shots is considerably higher than his usual performance. This Z-score is utilized to find the corresponding p-value and make decisions regarding the null hypothesis in hypothesis testing.