91Ó°ÊÓ

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Mistakes in a Song A random sample of six music students played a short song, and the number of mistakes in music each student made was recorded. After they practiced the song 5 times, the number of mistakes each student made was recorded. The data are shown. At \(\alpha=0.05,\) can it be concluded that there was a decrease in the mean number of mistakes? $$ \begin{array}{l|cccccc}{\text { Student }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} \\ \hline \text { Before } & {10} & {6} & {8} & {8} & {13} & {8} \\ \hline \text { After } & {4} & {2} & {2} & {7} & {8} & {9}\end{array} $$

Step by step solution

01

State the Hypotheses

We need to test if there is a decrease in the mean number of mistakes after practice. The null hypothesis (H0) states that there is no difference in the mean number of mistakes before and after practice. The alternative hypothesis (H1) states that the mean number of mistakes after practice is lower than before.Let \( \mu_d \) be the mean difference (before - after):- Null Hypothesis \( H_0: \mu_d = 0 \)- Alternative Hypothesis \( H_1: \mu_d > 0 \).The claim is that there was a decrease in the mean number of mistakes, which corresponds to the alternative hypothesis.

02

Find the Critical Value(s)

This is a one-tailed test at a significance level of \( \alpha = 0.05 \). Using the t-distribution table, for \( n-1 = 6-1 = 5 \) degrees of freedom, the critical t-value for a one-tailed test is approximately \( t_{critical} = 2.015 \).

03

Compute the Test Value

First, calculate the differences before and after for each student:\[\text{Differences: } [10-4, 6-2, 8-2, 8-7, 13-8, 8-9] = [6, 4, 6, 1, 5, -1]\]Next, calculate the mean (\( \bar{d} \)) and standard deviation (\( s_d \)) of these differences:- Mean \( \bar{d} = \frac{6 + 4 + 6 + 1 + 5 - 1}{6} = 3.5 \)- Standard deviation \( s_d = \sqrt{\frac{(6-3.5)^2 + (4-3.5)^2 + (6-3.5)^2 + (1-3.5)^2 + (5-3.5)^2 + (-1-3.5)^2}{5}} \approx 2.658 \)Now, compute the test statistic:\[t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{3.5}{2.658/\sqrt{6}} \approx 3.211\]

04

Make the Decision

Compare the calculated test statistic \( t = 3.211 \) with the critical value \( t_{critical} = 2.015 \). Since the test statistic is greater than the critical value, we reject the null hypothesis.

05

Summarize the Results

Since the null hypothesis is rejected, we conclude that there is sufficient evidence at the \( \alpha = 0.05 \) level of significance to support the claim that there is a decrease in the mean number of mistakes after the students practiced the song.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution

In hypothesis testing, the t-distribution is often used when dealing with small sample sizes, typically less than 30. It's a probability distribution that helps in estimating population parameters when the sample standard deviation is unknown, making it quite essential in t-tests.
The t-distribution is similar in shape to the normal distribution but has heavier tails. This means that it acknowledges that more extreme values are likely with a small sample, addressing the uncertainty more robustly compared to a normal distribution.
In our exercise, since we are dealing with a sample of six music students, the t-distribution is the appropriate choice to use in this scenario. Using it allows us to compute a more accurate critical value and test statistic for our hypothesis testing.

null hypothesis

The null hypothesis, denoted as H0, is an initial statement claiming there is no significant effect or difference in a particular condition or intervention. It serves as a starting assumption in hypothesis testing.
In the exercise, the null hypothesis states that there is no difference in the mean number of mistakes the students make before and after practicing. Mathematically, it is formulated as:

• \( H_0: \mu_d = 0 \)

This implies that the mean difference (\(\mu_d\)) between mistakes before and after practice is zero, suggesting practice has no impact. Testing the null hypothesis helps us determine if the observed differences are due to random chance or a true effect.

alternative hypothesis

When conducting hypothesis testing, the alternative hypothesis challenges the null hypothesis, proposing that there is indeed an effect or difference. It's what researchers often aim to prove true.
In this case, the alternative hypothesis, denoted as \(H_1\), posits that the mean number of mistakes made by the students after practicing is less than before. Formally, it is expressed as:

• \( H_1: \mu_d > 0 \)

This indicates that the mean difference (\(\mu_d\)) is greater than zero, suggesting the practice effectively reduced mistakes. The alternative hypothesis is often associated with the research claim we are trying to find evidence in favor of.

critical value

The critical value is a key threshold in hypothesis testing. It helps determine the boundary at which we decide whether a test statistic is extreme enough to reject the null hypothesis.
In a t-test, the critical value is taken from the t-distribution table based on the chosen significance level (\(\alpha\)), and the degrees of freedom (\(df\)). For our exercise, with \(\alpha = 0.05\) and \(df = 5\) (since there are 6 students), the critical value is approximately \(t_{critical} = 2.015\).
This critical value is compared to the test statistic. If the test statistic exceeds the critical value, it falls into the "rejection region," leading to the rejection of the null hypothesis in favor of the alternative.

test statistic

The test statistic is a standardized value computed from sample data during a hypothesis test. It helps determine how far the sample statistic is from the null hypothesis.In our context, the test statistic is calculated using the differences in the number of mistakes before and after practicing. First, the mean (\(\bar{d}\)) and standard deviation (\(s_d\)) of the differences are found. The formula used is:

• \( t = \frac{\bar{d}}{s_d/\sqrt{n}} \)

Plugging in our values gives \(t \approx 3.211\).
The test statistic of \(3.211\) is then compared to the critical value. With \(t\) being greater than \(t_{critical}\), we reject the null hypothesis, supporting that practice reduced mistakes.