Question

Heights of 9 -Year-Olds 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

No, the data do not support the claim that there is a difference in heights.

Step-by-step solution

Understanding the Hypotheses

The problem presents a claim test about the difference in average heights between boys and girls. We will utilize a two-sample t-test. The null hypothesis is that there is no difference in average heights: \(H_0: \mu_1 = \mu_2\). The alternative hypothesis is that there is a difference: \(H_a: \mu_1 eq \mu_2\).

Collecting Sample Data

From the data provided, we have:- Sample size for Boys \(n_1 = 60\), Mean height for Boys \(\bar{x}_1 = 123.5\) cm, Variance for Boys \(s_1^2 = 98\).- Sample size for Girls \(n_2 = 50\), Mean height for Girls \(\bar{x}_2 = 126.2\) cm, Variance for Girls \(s_2^2 = 120\).

Calculate the Test Statistic

We will use the formula for the t-statistic for two independent samples:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values:\[ t = \frac{123.5 - 126.2}{\sqrt{\frac{98}{60} + \frac{120}{50}}} \]

Compute the Numerator of the Test Statistic

Calculate the difference in means:\[ 123.5 - 126.2 = -2.7 \]

Compute the Denominator of the Test Statistic

Calculate each component under the square root:\[ \frac{98}{60} = 1.6333 \quad \text{and} \quad \frac{120}{50} = 2.4 \]Add these to get the total under the square root:\[ \sqrt{1.6333 + 2.4} = \sqrt{4.0333} = 2.008 \]

Calculate the t-Statistic

Now, divide the difference of means by the denominator:\[ t = \frac{-2.7}{2.008} \approx -1.344 \]

Determine the Degrees of Freedom and Critical Value

Use the formula for degrees of freedom for two samples:\[ df = \left( \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \right) \approx 107 \] Using a t-table, determine the critical value for \(\alpha = 0.05\) (two-tailed) is approximately \(\pm 1.984\).

Make a Decision

Since \(-1.344\) is within the range \(-1.984, 1.984\), we fail to reject the null hypothesis. Thus, there is not enough evidence to support the claim that there is a difference in heights at \(\alpha = 0.05\).

Key concepts

Two-sample t-test

The two-sample t-test is a statistical method used when you want to compare the means of two independent groups to see if there is a significant difference between them. For instance, in our provided exercise, we want to know if the height of 9-year-old boys is different from that of 9-year-old girls.

• Assumptions: Both groups have normal distributions, and the data samples are independent.
• Key Formula: The formula to calculate the t-statistic is \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

This formula helps in determining the statistical difference (if any) between the means of the two groups. In our exercise, the boys and girls each represent one of the groups. The calculated t-statistic helps us compare it against the critical t-value to decide if the difference is statistically significant.

Null hypothesis

The null hypothesis, often denoted as \(H_0\), is a statement of no effect or no difference. It's the default assumption that there’s no relationship between two variables or sets of data. In the context of the exercise, the null hypothesis is that there is no difference in average heights between 9-year-old boys and girls, which is expressed as \(H_0: \mu_1 = \mu_2\).

The null hypothesis serves a crucial role as it provides a baseline for testing the statistical evidence. We assess this hypothesis to see if the apparent results could happen by random chance. If our test statistic shows a significant difference, we may reject this hypothesis. Conversely, if not, we fail to reject it, implying no statistically significant difference was found.

Degrees of freedom

Degrees of freedom (df) are a concept that refers to the number of values in a calculation that can vary. They play a critical role in determining the shape of the t-distribution and are used in determining critical t-values from statistical tables. In the two-sample t-test, the degrees of freedom help adjust for the size of each sample and variance.

For our exercise, we use the formula:\[ df = \left( \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \right) \]
This formula results in an approximate df of 107. With this df, we can accurately assess how rare the calculated t-score would be under the null hypothesis. More degrees of freedom typically indicate better estimates of population parameters, resulting in more reliable statistical tests.

Critical value

The critical value is the threshold or cutoff point that helps decide whether to reject the null hypothesis. With a two-tailed test and a given significance level (like \(\alpha = 0.05\)), the critical value marks the boundaries of what constitutes a rare or unlikely event under the assumption that the null hypothesis is true.

To find the critical value for our exercise, first determine the degrees of freedom, which is approximately 107, and look up the critical values in a t-distribution table. For a two-tailed test at the 0.05 significance level, the critical values are approximately \(\pm 1.984\).

This means if the calculated t-statistic falls outside this range, it would be considered evidence enough to reject the null hypothesis. In this scenario, however, the t-statistic calculated was \(-1.344\), which does not fall outside these critical values. Thus, there is insufficient evidence to support the claim that the heights are different.