Question

Let \(p\) denote the probability that, for a particular tennis player, the first serve is good. Since \(p=0.40\), this player decided to take lessons in order to increase \(p\). When the lessons are completed, the hypothesis \(H_{0}: p=0.40\) will be tested against \(H_{1}: p>0.40\) based on \(n=25\) trials. Let \(y\) equal the number of first serves that are good, and let the critical region be defined by \(C=\\{y: y \geq 13\\}\). (a) Determine \(\alpha=P(Y \geq 13 ; ; p=0.40)\). (b) Find \(\beta=P(Y<13)\) when \(p=0.60\); that is, \(\beta=P(Y \leq 12 ; p=0.60)\) so that \(1-\beta\) is the power at \(p=0.60\).

Short answer

The calculated values of \(\alpha\) and \(\beta\) would be the short answer. The precise values would depend on the calculations made using the formulae provided above, and therefore cannot be stated without knowing the results of those calculations.

Step-by-step solution

Determine Alpha

To determine \(\alpha=P(Y \geq 13 ; ; p=0.40)\), calculate the probability of getting good serves 13 or more times out of 25 with a probability of successful serve being 0.40. We use a binomial distribution for this calculation. The formula for binomial distribution is \(P(X=k) = C(n, k)*(p^k)*((1-p)^(n-k))\) where n is the number of trials, p is the success probability per trial, and k is the number of successful trials. Because we want the probability for all outcomes from 13 to 25, we sum up the probabilities for each \(k\). Alternatively, we can calculate the complementary probability \(P(Y<13)\) and subtract it from 1, which could be more computationally efficient.

Calculate Alpha

Calculate \(1-P(Y<13 ; p=0.40)\) using binomial distribution formula summed up for \(k\) from 0 to 12. For each \(k\), calculate `C(25, k) * (0.4^k) * (0.6^(25-k))` and sum up the results. Subtract the result from 1 to get \(\alpha\). Use a calculator or a software that can handle binomial calculations to get the numerical value of \(\alpha\).

Determine Beta

To find \(\beta=P(Y<13; p=0.60)\), calculate the probability of getting good serves 12 times or less out of 25 with a success probability per serve of 0.6, again using a binomial distribution. The calculation process is similar to that of determining \(\alpha\), just with different parameters (\(k\) from 0 to 12 and \(p=0.60\)).

Calculate Beta

Calculate \(\beta=P(Y<13, p=0.60)\) using binomial distribution formula summed up for \(k\) from 0 to 12. For each \(k\), calculate `C(25, k) * (0.6^k) * (0.4^(25-k))` and sum up the results. Use a calculator or software with binomial calculation function to get a numerical value for \(\beta\).

Key concepts

Binomial Distribution

The binomial distribution is a fundamental concept in statistics that describes the number of successes in a fixed number of independent trials, with each trial having the same probability of success. This distribution is defined by two parameters: the number of trials (denoted as \( n \)) and the probability of success for each trial (denoted as \( p \)).

For example, if a tennis player attempts 25 first serves, each with an independent chance of being good (a 'success'), and the probability of each successful serve is \( p \), the scenario can be modeled by a binomial distribution. The probability of having exactly \( k \) good serves is given by the formula:
\[ P(X=k) = C(n, k) \times (p^k) \times ((1-p)^{n-k}) \]
Here, \( C(n, k) \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials. To find the probability of having at least a certain number of successes, one would sum the probabilities from that number up to \( n \), or alternatively, subtract the cumulative probability of having fewer successes from 1.

Probability of Success

In our exercise, the probability of success, \( p \), represents the chance that the tennis player's first serve is good. Originally, the player's probability of a successful first serve is \( p=0.40 \). After taking lessons, there's a hypothesis that this probability has increased, which means \( p \) should be greater than 0.40.

Understanding \( p \) is crucial for performing hypothesis testing. When you conduct a test, you're essentially evaluating whether the observed data (the number of good serves) align with the expected probability of success or if they suggest a different probability. An important application of \( p \) is to determine the expected value and variance of the binomial distribution, which are \( np \) and \( np(1-p) \), respectively.

Critical Region

A critical region is a set of outcomes of an experiment that leads to the rejection of the null hypothesis, \( H_0 \). It is determined based on a chosen significance level, \( \alpha \), which is the probability of making a Type I error—rejecting the null hypothesis when it's in fact true.

In our tennis player's context, the critical region is defined as \( C=\{y: y \geq 13\} \). This means that if the player makes 13 or more successful serves out of 25 after taking lessons, we will reject the hypothesis that the player’s probability of a good first serve remains unchanged at \( p=0.40 \). The choice of the critical region affects the test's ability to detect an actual increase in the probability of a successful first serve, a concept closely tied to the power of a test.

Power of a Test

The power of a test, denoted as \( 1-\beta \), refers to the probability that a test correctly rejects a false null hypothesis. In our exercise, we seek to determine the power of the test when the true probability of a successful serve, \( p \), has increased to 0.60 after the tennis player has completed lessons.

The power is calculated by finding \( \beta=P(Y<13; p=0.60) \), the probability of observing fewer than 13 successful serves (the complement of our critical region) when the true success probability is 0.60. The power is then \( 1-\beta \), representing the chance that the test successfully detects the improvement in the player's serve. High power is desirable as it means the test is sensitive and can reliably identify true effects. For meaningful results, statisticians usually aim for a test power of at least 0.80, or 80%.