Question

To determine whether there is a significant difference in the average weights of boys and girls beginning kindergarten, random samples of 50 5-year-old boys and 50 5-year-old girls produced the following information. \(^{17}\) $$\begin{array}{llcc}\hline & & \text { Standard } & \text { Sample } \\\& \text { Mean } & \text { Deviation } & \text { Size } \\\\\hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\\\\text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \\\\\hline\end{array}$$ Find a \(99 \%\) lower confidence bound for the difference in the average weights of 5 -year-old boys and girls. Can you conclude that on average 5 -year-old boys weigh more than 5-year-old girls?

Short answer

Answer: Yes, based on the analysis, we can conclude that on average, 5-year-old boys weigh more than 5-year-old girls at a 99% confidence level.

Step-by-step solution

Calculate the Standard Error

The first step is to calculate the standard error for the difference between the sample means. The formula for the standard error (SE) of the difference between two sample means is: $$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ Where \(s_1\) and \(s_2\) are the standard deviations of the two samples and \(n_1\) and \(n_2\) are the respective sample sizes. Plugging in the given values, we get: $$SE = \sqrt{\frac{2.4^2}{50} + \frac{1.9^2}{50}} = \sqrt{\frac{5.76}{50} + \frac{3.61}{50}} = \sqrt{0.1872} \approx 0.43$$

Calculate the Degrees of Freedom

The next step is to calculate the degrees of freedom for the t-distribution. This can be calculated using the following formula: $$df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}$$ Plugging in the values, we get: $$df = \frac{(0.1872)^2}{\frac{(0.1152)^2}{49} + \frac{(0.0722)^2}{49}} \approx 85.7$$ We will round it down to 85.

Find the Critical Value

Now we need to find the critical value from the t-distribution table. We are looking for a 99% lower confidence bound, so we will calculate the critical value (\(t^*\)) at the 1% significance level with 85 degrees of freedom. Looking up in a t-distribution table or using a calculator, \(t^* \approx 2.636\).

Calculate the Lower Confidence Bound

Now, we have all the information required to calculate the lower confidence bound for the difference in average weights of boys and girls. The formula for the lower confidence bound (\(L\)) for the difference between two sample means is: $$L = (\bar{x}_1 - \bar{x}_2) - t^* \times SE$$ Plugging in the sample means and the values we calculated earlier, we get: $$L = (19.4 - 17.0) - 2.636 \times 0.43 \approx 1.067$$

Conclusion

The 99% lower confidence bound for the difference in the average weights of 5-year-old boys and girls is approximately 1.067 kg. Since the value is positive, we can conclude that 5-year-old boys on average weigh more than 5-year-old girls, as the lower bound at 99% confidence level isn't 0 or negative.

Key concepts

Standard Error Calculation

Understanding the standard error (SE) is critical in statistics as it signifies the variability or precision of an estimate from a sample. When comparing two sample means, like the average weights of boys and girls, the standard error of the difference between those means helps assess how much the sample means might differ due to random chance alone.

In the given problem, the SE is calculated using the formula:
\[\begin{equation}SE = \[\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\frac{\]\]\end{equation}\]

where
- \( s_1 \) and \( s_2 \) represent the sample standard deviations, which are measures of the spread of scores within each group of boys and girls, respectively.
- \( n_1 \) and \( n_2 \) refer to the sample sizes, which in both cases are 50.

By plugging in the values, we found the SE to be approximately 0.43, which reflects the extent to which the difference between boys' and girls' average weights might vary if we were to take multiple samples from the population.

Degrees of Freedom

The concept of degrees of freedom (df) relates to the number of independent values or quantities which can be assigned to a statistical distribution. The df can be thought of as the amount of 'wiggle room' available when estimating certain parameters from samples of a population. In the context of a t-distribution, the degrees of freedom are an essential aspect because they determine the shape of the distribution curve and the critical t-values associated with it.

In our exercise, the df for the problem was calculated using a complex formula:
\[\begin{equation}df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}\end{equation}\]

This formula incorporates both sample sizes and variances, resulting in a dof value of about 85.7, which was rounded down to 85. The df not only informs the shape of our t-distribution but also has a crucial role in determining how much the sampled data can vary, considering there is a fixed total sample size.

T-Distribution

The t-distribution, also known as Student's t-distribution, is a probability distribution that is symmetrical and bell-shaped, like the standard normal distribution, but it has heavier tails. This means there is a greater likelihood of finding values at the extremes, and it reflects the added uncertainty due to estimating the population standard deviation from the sample.

In the case of comparing two small sample means, where the population standard deviation is unknown, the t-distribution is used to derive critical values for confidence intervals and hypothesis tests. The t-distribution varies depending on the degrees of freedom; as the df increases, it closely resembles the standard normal distribution.

For our exercise, using the t-distribution with 85 df, we obtained a critical value (( t^* )) of approximately 2.636 at the 1% significance level. This value helps us define the lower boundary of the confidence interval for the mean difference between weights of boys and girls, ensuring the conclusion drawn from the sample is not merely by chance.