Question

Independent random samples were selected from two quantitative populations, with sample sizes, means, and variances given in Exercises 13-14. Construct a 95\% upper confidence bound for \(\mu_{1}-\mu_{2}\). Can you conclude that one mean is larger than the other? If so, which mean is larger? $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 64 & 64 \\\\\text { Sample Mean } & 3.9 & 5.1 \\\\\text { Sample Variance } & 9.83 & 12.67 \\\\\hline\end{array}$$

Short answer

Explain your answer.

Step-by-step solution

Calculate the sample standard deviation

For each population, we need to find the sample standard deviation (s1 and s2) using the sample variance. The standard deviation can be found by taking the square root of the variance: $$s_1 = \sqrt{9.83} = 3.135$$ $$s_2 = \sqrt{12.67} = 3.560$$

Calculate the standard error of the difference

To calculate the standard error of the mean difference, use the given sample sizes (n1 and n2) and the standard deviations (s1 and s2) found in Step 1: $$SE(\bar{X}_1 - \bar{X}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3.135^2}{64} + \frac{3.560^2}{64}} = 0.576$$

Find the degrees of freedom

To find the t-distribution's degrees of freedom, use the following formula: $$df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2 / n_1)^2}{n_1 - 1} + \frac{(s_2^2 / n_2)^2}{n_2 - 1}} = \frac{(0.576)^2}{\frac{(3.135^2 / 64)^2}{63} + \frac{(3.560^2 / 64)^2}{63}} = 126.097$$ The degrees of freedom are approximately 126.

Find the critical t-value

To find the critical t-value, we need the 95% upper confidence bound (or 0.025 in one-tail) and the degrees of freedom (126). Using a t-table or a calculator, we find that the critical t-value is: $$t_{0.025,126} = 1.980$$

Calculate the upper confidence bound

Use the critical t-value, sample means, and standard error to calculate the 95% upper confidence bound for the difference between the population means (\(\mu_{1} - \mu_{2}\)): $$\text{Upper Confidence Bound} = (\bar{X}_1 - \bar{X}_2) + t_{0.025,126} \times SE(\bar{X}_1 - \bar{X}_2)$$ $$= (3.9 - 5.1) + 1.980 \times 0.576 = -1.2 + 1.141 = -0.059$$

Determine if one mean is larger than the other

Since the 95% upper confidence bound for \(\mu_{1} - \mu_{2}\) is -0.059, which lies below zero, we can conclude that \(\mu_{1}\) is not larger than \(\mu_{2}\). In fact, we can confidently state that \(\mu_{2}\) is larger than \(\mu_{1}\) at the 95% confidence level.

Key concepts

Confidence Interval

A confidence interval provides a range of values within which there's a specified probability that the value of a parameter lies. It's a way of expressing uncertainty in an estimate. When constructing a 95% upper confidence bound for the difference between two means, \( \mu_1 - \mu_2 \), as in the exercise, the upper bound represents the upper limit of the confidence interval, suggesting that there's a 95% chance the true difference between the means is less than this value.

Using this confidence interval, one can determine the likelihood that one population mean is greater than the other. If the upper confidence bound is above zero, \( \mu_1 \) may be greater; if below zero, as in the given exercise, we can be confident \( \mu_2 \) is larger. The calculation involves several steps including finding the standard error, degrees of freedom, and the appropriate critical value from the t-distribution.

Standard Error

In statistics, the standard error measures the accuracy with which a sample represents a population. In the context of the exercise, the standard error of the difference between two sample means \( SE(\overline{X}_1 - \overline{X}_2) \) gives an idea of how far the calculated difference between sample means might be from the true difference between population means.

This value is essential in constructing the confidence interval. It's calculated using the standard deviations and sample sizes from both samples. The smaller the standard error, the more precise the estimate is likely to be. In the provided solution, the standard error plays a crucial role in calculating the upper bound of the confidence interval.

Degrees of Freedom

Degrees of freedom are a vital concept in the realm of statistics, referring to the number of independent values in a calculation that are free to vary. When estimating the variance from the sample, the degrees of freedom equate to the sample size minus one. For a t-distribution, which is used when the population variance is unknown and the sample size is relatively small, the degrees of freedom help in determining the appropriate t-statistic to use.

In the solution steps, we see the use of a complex formula to calculate the degrees of freedom. This accounts for the fact that we're dealing with two different samples and is essential for correctly identifying the critical t-value when finding the upper confidence bound.

T-distribution

The t-distribution, or Student's t-distribution, is a probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown. It's similar to the normal distribution but has thicker tails, meaning it's more prone to producing values that fall far from its mean.

This is crucial to the exercise because it acknowledges the additional uncertainty we face due to estimating the population standard deviation from our samples. With the degrees of freedom from the previous steps, one can find the critical t-value which determines the multiplier for the standard error in the confidence interval calculation. As the sample size grows larger, the t-distribution approaches the normal distribution.