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. At age 9 the average weight \((21.3 \mathrm{~kg})\) and the average height \((124.5 \mathrm{~cm})\) for both boys and girls are exactly the same. A random sample of 9-year-olds yielded these results. At \(\alpha=0.05,\) do the data support the given claim that there is a difference in heights? $$ \begin{array}{lcc} & \text { Boys } & \text { Girls } \\ \hline \text { Sample size } & 60 & 50 \\ \text { Mean height, } \mathrm{cm} & 123.5 & 126.2 \\ \text { Population variance } & 98 & 120 \end{array} $$

Short answer

The data do not support the claim that there is a difference in heights at \( \alpha = 0.05 \).

Step-by-step solution

State the Hypotheses and Identify the Claim

We need to determine if there is a significant difference in heights between boys and girls. The null hypothesis \( H_0 \) states that there is no difference in the mean heights: \( \mu_{boys} = \mu_{girls} \). The alternative hypothesis \( H_1 \) states that there is a difference: \( \mu_{boys} eq \mu_{girls} \). This makes it a two-tailed test.

Find the Critical Values

For a two-tailed test with \( \alpha = 0.05 \), we find critical values for a normal distribution. Using a standard Z-table, the critical z-values are approximately \( \pm 1.96 \). Thus, our critical region will be any z-value less than -1.96 or greater than 1.96.

Calculate the Test Statistic

Since the population variances are known, we can use the formula for the z-test statistic for two means:\[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Where \( \bar{x}_1 = 123.5 \), \( \bar{x}_2 = 126.2 \), \( \sigma_1^2 = 98 \), \( \sigma_2^2 = 120 \), \( n_1 = 60 \), and \( n_2 = 50 \). Substitute the values to compute:\[ z = \frac{(123.5 - 126.2)}{\sqrt{\frac{98}{60} + \frac{120}{50}}} \approx \frac{-2.7}{\sqrt{1.6333 + 2.4}} \approx \frac{-2.7}{\sqrt{4.0333}} \approx \frac{-2.7}{2.0083} \approx -1.34 \]

Make the Decision

The calculated test statistic is \( z = -1.34 \). Compare this with the critical z-values \( \pm 1.96 \). Since \(-1.34\) is greater than \(-1.96\) and less than \(1.96\), it falls within the range of not rejecting the null hypothesis. Therefore, we do not have enough evidence to reject \( H_0 \).

Summarize the Results

At the level of significance \( \alpha = 0.05 \), the data do not support the claim that there is a difference in the mean heights of 9-year-old boys and girls since the test statistic \(-1.34\) does not fall in the critical region. There is insufficient evidence to suggest a significant difference in heights.

Key concepts

Two-Tailed Test

When dealing with statistics and hypothesis testing, a two-tailed test is crucial to understand. It allows us to explore the possibility of a relationship or difference in either direction. Unlike a one-tailed test, where we are only looking for either an increase or a decrease, a two-tailed test checks for any significant difference.

In the context of comparing the heights of boys and girls at age 9, the two-tailed test evaluates whether the average height differs on either side: taller or shorter than each other. This kind of test is symmetric, considering deviations both higher and lower than the mean difference specified. For the given problem, the null hypothesis states that there is no difference in heights, while the alternative hypothesis posits a noticeable difference, either boys being taller than girls or the other way around.

Two-tailed tests are particularly useful when uncertainty exists about the direction of the difference, making them a common choice in many analytical scenarios.

Z-Test Statistic

The z-test statistic is central to determining whether the difference between sample means is statistically significant. It assumes that the data follows a normal distribution or that each sample size is large enough to employ the central limit theorem effectively.

In the exercise, we're comparing sample means of heights using known population variances. The z-test formula is:

• \( z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \)

Here, \( \bar{X}_1 \) and \( \bar{X}_2 \) are the sample means for boys and girls respectively, \( \sigma_1^2 \) and \( \sigma_2^2 \) the known population variances, and \( n_1 \), \( n_2 \) the sample sizes.

In our problem, the calculated z-value was approximately -1.34, indicating the position of our result relative to the hypothesized mean difference. The z-value helps us ascertain how many standard deviations away our observed statistic lies from the expected, under the assumption of the null hypothesis.

Critical Values

Critical values are the thresholds set to determine whether to reject a null hypothesis. They define the boundaries of the critical region where the observed test statistic must lie for us to consider it statistically significant.

In hypothesis testing, particularly in a two-tailed test, we use two critical values which correspond to the edges of the distribution tails, especially for the standard normal distribution. For an alpha level \( \alpha = 0.05 \), these critical z-values typically fall at \( \pm 1.96 \). This means any test statistic further away from the mean than these critical values indicate significance beyond the level of risk (\( \alpha \)) we're willing to accept.

For our exercise, the computed test statistic of approximately -1.34 lies between -1.96 and 1.96, hence not in the critical region, leading us to retain the null hypothesis and conclude that there is not enough evidence to show a significant difference in heights.

Null and Alternative Hypotheses

The null and alternative hypotheses are the backbone of hypothesis testing. They establish the basis for testing any claim in statistical analysis.

The null hypothesis, denoted as \( H_0 \), suggests that there is no effect or difference. In our exercise, the null hypothesis stated that the mean height of boys equals that of girls: \( \mu_{boys} = \mu_{girls} \). The alternative hypothesis, represented as \( H_1 \), is what we are testing to determine: whether a significant difference actually exists, \( \mu_{boys} eq \mu_{girls} \).

The aim is to see if we have sufficient statistical evidence to reject the null hypothesis in favor of the alternative. In scenarios where the calculated z-test statistic falls outside the range of critical values, we conclude a statistical significance favoring \( H_1 \). Understanding this concept is crucial, as it drives the direction and decisions in statistical hypothesis testing across various fields.