Question

Perform each of the following steps. 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. A random survey of 1000 students nationwide showed a mean ACT score of 21.4. Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is 3 , can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

Short answer

Ohio's mean ACT score is significantly below the national average at α=0.05.

Step-by-step solution

State the Hypotheses

First, we need to state our null and alternative hypotheses. We want to test if the mean ACT score for Ohio students is below the national average. - Null Hypothesis ( H_0 ): μ = 21.4 (The mean ACT score for Ohio is equal to the national average). - Alternative Hypothesis ( H_1 ): μ < 21.4 (The mean ACT score for Ohio is below the national average). The claim is in the alternative hypothesis, aiming to show that Ohio is below the national average.

Find the Critical Value

Given that α = 0.05 and this is a left-tailed test (since we are testing if Ohio is below the average), we need to find the critical value of z for α = 0.05 . Using a standard normal distribution table, the critical value (z_{ ext{critical}}) for α = 0.05 in a left-tailed test is approximately -1.645 .

Compute the Test Value

We use the formula for the z ext{-test value}:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]where:- \bar{x} = 20.8 (Ohio's sample mean)- \mu = 21.4 (national average mean)- \sigma = 3 (population standard deviation)- n = 500 (sample size for Ohio)Calculate:\[z = \frac{20.8 - 21.4}{\frac{3}{\sqrt{500}}} = \frac{-0.6}{\frac{3}{22.36}} = \frac{-0.6}{0.134} \approx -4.48\]

Make the Decision

In this step, we compare the calculated test value with the critical value. Since -4.48 is less than -1.645 , we fall within the rejection region. Therefore, we reject the null hypothesis, H_0 .

Summarize the Results

We have sufficient evidence to support the claim that the mean ACT score for Ohio students is below the national average at the α = 0.05 significance level. This suggests that Ohio's students, on average, have a lower ACT score compared to the national average.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement we assume to be true until we have enough evidence to suggest otherwise. It often represents a status quo or a default position. In our exercise with ACT scores, the null hypothesis ( H_0 ) states that the mean ACT score for Ohio students is equal to the national average, specifically 21.4.

Formulating this hypothesis allows us to have a starting point for statistical testing. If we gather sufficient data that contradicts the null hypothesis, we can make the decision to reject it in favor of an alternative explanation.

• The null hypothesis is generally expressed with an equals sign. For our case, it is: H_0: μ = 21.4 , where μ represents the population mean ACT score for Ohio.
• We do not initially assume that the null hypothesis is false; rather, we seek evidence to potentially refute it.

Alternative Hypothesis

The alternative hypothesis is what we consider when the null hypothesis is rejected. It's effectively the claim we aim to prove, providing a different perspective from the null. In our example of evaluating Ohio's ACT scores, the alternative hypothesis ( H_1 ) suggests that the mean ACT score for Ohio students is less than the national average, hence μ < 21.4.

This hypothesis insinuates a potential deviation from the national average, aligning with our curiosity to determine whether Ohio's students perform below the national norm.

• The alternative hypothesis uses inequality signs, capturing our research interest, which in this scenario is: H_1: μ < 21.4 .
• It's key to note that the alternative hypothesis is the one we're typically vested in proving with statistical evidence.
• Deciding for this hypothesis often concludes that there is statistically significant evidence, suggesting a difference or effect.

Critical Value

The critical value is a threshold that helps in making a decision to reject the null hypothesis or not. It corresponds to a particular probability and the significance level ( α ) we choose for our test. In hypothesis testing, determining this value is crucial, as it defines the cutoff for the rejection region.

For our exercise, with a significance level of 0.05 in a left-tailed test, the critical value from the standard normal distribution was found to be approximately -1.645 . This is the boundary at which we decide whether or not the observed data lead to rejecting the null hypothesis.

• The critical value is dependent on the chosen significance level ( α ). In our case, α = 0.05 setting a boundary for our acceptable evidence level.
• It helps demarcate the edge of the rejection region, offering a visual boundary in statistical diagrams.
• When the calculated test statistic falls beyond this value towards the left, it reinforces our rejection of the null hypothesis.

Z-Test

A Z-test is a statistical method we use when we want to test hypotheses about a population mean, with data assumed to follow a normal distribution. Specifically, it's designed for situations with known variances and a sufficiently large sample size, which matches the conditions of our ACT score analysis.

The formula used in a Z-test is: \[z = \frac{\bar{x} - μ}{\frac{σ}{\sqrt{n}}}\]where:

• \( \bar{x} \) is the sample mean, which is 20.8 in our Ohio ACT scores context.
• \( μ \) is the population mean, or national average, set at 21.4.
• \( σ \) is the known population standard deviation of 3.
• \( n \) is the sample size, 500 for Ohio students.

Using these values, we calculated a Z-test statistic of approximately -4.48. When this statistic is compared to the critical value, we conclude if the null hypothesis holds or if the alternative takes precedence.

Significance Level

The significance level ( α ) in hypothesis testing conveys the probability of making a Type I error: rejecting the true null hypothesis. It's the potential risk we are prepared to accept in making our statistical decision. In our problem dealing with ACT scores, we had set α = 0.05 , which is a conventionally accepted level in many scientific inquiries.

This ratio indicates our tolerance for incorrectly declaring Ohio students' ACT performance as significantly below average when it might not truly differ.

• The significance level quantifies the chance of taking a false positive action, suggesting a moderated move towards cautious conclusions.
• Common levels include 0.05 , 0.01 , and 0.10 , depending heavily on the context and field of study.
• A lower α makes the rejection of a null hypothesis more stringent, suggesting cautiousness in asserting statistical significance.

By controlling this level, researchers balance their confidence in results against the likelihood of spurious conclusions.