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. The average monthly Social Security benefit for a specific year for retired workers was \(\$ 954.90\) and for disabled workers was \(\$ 894.10 .\) Researchers used data from the Social Security records to test the claim that the difference in monthly benefits between the two groups was greater than \(\$ 30 .\) Based on the following information, can the researchers' claim be supported at the 0.05 level of significance? $$ \begin{array}{lll} & \text { Retired } & \text { Disabled } \\ \hline \text { Sample size } & 60 & 60 \\ \text { Mean benefit } & \$ 960.50 & \$ 902.89 \\ \text { Population standard deviation } & \$ 98 & \$ 101 \end{array} $$

Short answer

The claim that the difference is greater than $30 cannot be supported at the 0.05 significance level.

Step-by-step solution

State the Hypotheses and Identify the Claim

We need to formulate our null and alternative hypotheses. The claim is that the difference in monthly benefits between retired and disabled workers is greater than $30.- Null Hypothesis \(H_0\): \(\mu_1 - \mu_2 \leq 30\) - Alternative Hypothesis \(H_a\): \(\mu_1 - \mu_2 > 30\)Here, \(\mu_1\) represents the mean benefit for retired workers, and \(\mu_2\) represents the mean benefit for disabled workers. The claim \(\mu_1 - \mu_2 > 30\) is the alternative hypothesis.

Find the Critical Value(s)

With a significance level \(\alpha = 0.05\) and a one-tailed test because the alternative hypothesis includes \(> 30\), we need to find the critical value for the standard normal distribution (Z-distribution). For a one-tailed test at \( 0.05 \) level of significance, the critical value \(Z_c\) is approximately 1.645, which corresponds to the 95th percentile.

Compute the Test Value

To calculate the test value, we use the formula for the Z-test for two means:\[Z = \frac{(\bar{X}_1 - \bar{X}_2) - D}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Where:- \(\bar{X}_1 = 960.50\)- \(\bar{X}_2 = 902.89\)- \(D = 30\) (the hypothesized difference)- \(\sigma_1 = 98\), \(\sigma_2 = 101\)- \(n_1 = 60\), \(n_2 = 60\)Substituting the values:\[Z = \frac{(960.50 - 902.89) - 30}{\sqrt{\frac{98^2}{60} + \frac{101^2}{60}}} = \frac{57.61 - 30}{\sqrt{\frac{9604}{60} + \frac{10201}{60}}}\]\[Z = \frac{27.61}{\sqrt{160.0667 + 170.0167}} = \frac{27.61}{18.41} \approx 1.50\]

Make the Decision

Compare the calculated test value \(Z = 1.50\) with the critical value \(Z_c = 1.645\). Since \(1.50 < 1.645\), we fail to reject the null hypothesis.

Summarize the Results

At the 0.05 level of significance, we do not have sufficient evidence to support the claim that the difference in monthly benefits between retired and disabled workers is greater than $30. The data does not provide strong enough evidence that \(\mu_1 - \mu_2 > 30\).

Key concepts

Null Hypothesis

In hypothesis testing, the **null hypothesis** plays the pivotal role of setting a baseline assumption. It is like the status quo in an experiment, declaring that there is no effect or no difference between groups or conditions being studied. Specifically, in this exercise, the null hypothesis \(H_0\) suggests that the difference in mean monthly benefits between retired and disabled workers is less than or equal to $30. This can be mathematically expressed as \( \mu_1 - \mu_2 \leq 30.\)

This hypothesis assumes that any observed difference might just be due to random chance rather than a true effect. In statistical tests, our primary job is to challenge this null hypothesis to see if there is enough evidence to reject it. This does not mean that the null hypothesis is the statement ":there is no difference"; rather, it often includes a range of plausible values, as it does in this case where it includes \( \leq 30\) difference.

The null hypothesis is always tested with the intent to refute it in favor of an alternative hypothesis, unless evidence strongly suggests otherwise.

Alternative Hypothesis

The **alternative hypothesis** represents what a researcher aims to support through evidence. It stands in contrast to the null hypothesis and claims that there is a real effect or difference beyond random chance.

In this scenario, the alternative hypothesis \(H_a\) states that the difference in mean monthly benefits between retired and disabled workers is greater than \(30, i.e., \(\mu_1 - \mu_2 > 30\).

This hypothesis is central to the claim the researchers are testing. They are trying to demonstrate that the observed data provide sufficient evidence to show that the disparity in benefits exceeds \)30. The alternative hypothesis is what drives the research and hypothesis testing—researchers seek significant results to support this hypothesis.

We use the alternative hypothesis to guide the direction of the statistical test. In this exercise, it clearly suggests a one-tailed test since the claim is directional, specifying ">" rather than being non-directional ("not equal to").

Significance Level

The **significance level** \( \alpha \) is a threshold set by researchers before conducting a hypothesis test, which determines how strong the evidence must be before we can reject the null hypothesis. It reflects the probability of making a Type I error, which occurs when the null hypothesis is rejected when it is actually true.

**Setting the Scene with \( \alpha = 0.05 \):**

• An \( \alpha \) of 0.05 is commonly used and signifies a 5% risk of making a Type I error.
• If a test results in a p-value less than 0.05, the findings are typically considered statistically significant.
• This threshold helps researchers decide whether to reject the null hypothesis.

In this exercise, the significance level of 0.05 was chosen, which means the researchers are willing to accept a 5% chance of incorrectly asserting that the difference in monthly benefits exceeds $30.

The choice of significance level depends on the context of the hypothesis test. In some scientific fields, even more stringent levels like 0.01 are used to lower the risk of error.

Z-Test

A **z-test** is a statistical procedure that measures whether there is a significant difference between the means of two groups, using the standard normal distribution to determine how far away the sample means are from each other or from a hypothesized value.

In hypothesis testing, a z-test is appropriate when:

• The sample size is large enough, typically greater than 30.
• The population standard deviation is known.

For this particular exercise, the researchers applied a z-test formula to calculate a test statistic. This value helps us evaluate how many standard deviations the observed difference is from the hypothesized difference.

**Computing the Test Value:**The test statistic in z-tests for two means is given by:\[Z = \frac{(\bar{X}_1 - \bar{X}_2) - D}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]This formula provides a standardized way to assess whether the observed data supports the alternative hypothesis. Under the null hypothesis, this statistic follows a standard normal distribution.

Critical Value

The **critical value** is the cutoff point that defines the boundary for rejecting the null hypothesis. It is derived from the distribution that the test statistic follows—in this instance, the standard normal distribution.

When performing a hypothesis test, the critical value dictates whether the test statistic falls in the region where the null hypothesis is rejected. For a one-tailed z-test at a 0.05 significance level, the critical value is approximately 1.645. This means if the calculated test statistic exceeds 1.645, the null hypothesis can be rejected in favor of the alternative.

**What the Critical Value Tells Us:**

• If the test statistic is greater than the critical value in a right-tailed test, we reject the null hypothesis.
• The region beyond the critical value is known as the "critical region."

In this specific exercise, the computed test statistic was 1.50, which is less than the critical value of 1.645. Thus, the decision was to not reject the null hypothesis because the test statistic does not enter the critical region. This shows how the critical value is an essential tool in making formal decisions about hypotheses.