Question

Two independent random samples of sizes \(n_{1}=4\) and \(n_{2}=5\) are selected from each of two normal populations: $$ \begin{array}{l|ccccc} \text { Population } 1 & 12 & 3 & 8 & 5 \\ \hline \text { Population } 2 & 14 & 7 & 7 & 9 & 6 \end{array} $$ a. Calculate \(s^{2},\) the pooled estimator of \(\sigma^{2}\). b. Find a \(90 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right),\) the difference between the two population means. c. Test \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right)<0\) for \(\alpha=.05 .\) State your conclusions.

Short answer

Answer: The confidence interval for the difference between the population means is (-4.044, 0.844). The result of the hypothesis test indicates that we fail to reject the null hypothesis, meaning there is insufficient evidence to conclude that the difference in means is less than zero.

Step-by-step solution

Calculate Sample Means

First, we need to calculate the sample means for the two populations: $$\bar{X}_{1}=\frac{12+3+8+5}{4}=7$$ $$\bar{X}_{2}=\frac{14+7+7+9+6}{5}=8.6$$

Calculate Sample Variances

Next, we find the sample variances for each population: $$s_{1}^{2}=\frac{\sum_{i=1}^{4}\left(X_{1i}-\bar{X}_{1}\right)^{2}}{n_{1}-1}=\frac{(12-7)^2+(3-7)^2+(8-7)^2+(5-7)^2}{4-1}=\frac{49}{3}$$ $$s_{2}^{2}=\frac{\sum_{i=1}^{5}\left(X_{2i}-\bar{X}_{2}\right)^{2}}{n_{2}-1}=\frac{(14-8.6)^2+(7-8.6)^2+(7-8.6)^2+(9-8.6)^2+(6-8.6)^2}{5-1}=8.3$$

Calculate Pooled Variance Estimator

Now, we can calculate the pooled variance estimator: $$s^{2}=\frac{\left(n_{1}-1\right)s_{1}^{2}+\left(n_{2}-1\right)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(4-1)\frac{49}{3}+(5-1)8.3}{4+5-2}=6.514$$

Find the t-value for Confidence Interval

We want a \(90 \%\) confidence interval, which means \(\alpha\) (the level of significance) is \(0.1\). Since we have a two-tailed test, we will use \(\alpha/2=0.05\). We also need to use the degrees of freedom: \(\text{df}=n_{1}+n_{2}-2=4+5-2=7\). Using a t-distribution table or calculator, we find the t-value for a \(90 \%\) confidence interval: $$t_{\alpha/2,df}=t_{0.05,7}=1.895$$

Calculate Confidence Interval

Now, we will use the t-value and the pooled variance estimator to calculate the confidence interval for the difference in population means: $$\left(\mu_{1}-\mu_{2}\right)=\left(\bar{X}_{1}-\bar{X}_{2}\right)\pm t_{\alpha/2,df}\cdot\sqrt{\frac{s^2}{n_{1}}+\frac{s^2}{n_{2}}}$$ $$=(-1.6)\pm 1.895\cdot\sqrt{\frac{6.514}{4}+\frac{6.514}{5}}=(-1.6)\pm1.895\cdot 1.289=(-1.6)\pm2.444$$ The confidence interval for the difference between the population means is: $$(-4.044, 0.844)$$

Test Hypothesis

To test the null hypothesis \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against the alternative hypothesis \(H_{a}:\left(\mu_{1}-\mu_{2}\right)<0\), we calculate the t-value for our sample data: $$t=\frac{\left(\bar{X}_{1}-\bar{X}_{2}\right)-0}{\sqrt{\frac{s^2}{n_{1}}+\frac{s^2}{n_{2}}}}=\frac{-1.6}{1.289}=-1.241$$ We now compare our calculated t-value of \(-1.241\) to the critical t-value of \(-1.895\) (with 7 degrees of freedom and a one-tailed significance level of \(\alpha=0.05\)). Since \(-1.241 > -1.895\), we fail to reject the null hypothesis.

Conclusion

The confidence interval for the difference between the two population means is \((-4.044, 0.844)\). Based on our hypothesis test, we fail to reject the null hypothesis, which means there's insufficient evidence to conclude that the difference in means is less than zero.

Key concepts

Hypothesis Testing

Hypothesis testing is a fundamental part of statistical inference that allows us to make decisions or draw conclusions about a population based on sample data. In this context, we want to determine whether the difference between two population means is equal, or if one is less than the other.

For the given exercise, our null hypothesis (\(H_{0}\)) states that the difference between the population means is zero: \((\mu_{1}-\mu_{2})=0\). The alternative hypothesis (\(H_{a}\)) proposes that the difference is less than zero: \((\mu_{1}-\mu_{2})<0\).

We calculate a test statistic, which in this case is a t-value, based on our sample data. We then compare this value to a critical t-value from statistical tables, which helps us decide whether to reject or fail to reject the null hypothesis. If our calculated t-value is more extreme than the critical value, we reject \(H_{0}\). In this example, since \(-1.241 > -1.895\), we fail to reject the null hypothesis, meaning there's no strong evidence against it.

This process essentially allows us to weigh the evidence from the data against a pre-specified level of confidence (significance level \(\alpha\)). For a more robust grasp, remember always: hypothesis testing consists of assuming no effect or difference until evidence suggests otherwise.

Confidence Interval

A confidence interval provides a range of values that likely contain the true difference between the population means. This statistical tool helps us to understand the precision of our sample estimate.

In the exercise, we calculate a 90% confidence interval for the difference \((\mu_{1}-\mu_{2})\). The interval is computed using the sample means, the pooled variance, and a critical t-value, together forming the expression: \[ \left(\bar{X}_{1}-\bar{X}_{2}\right) \pm t_{\alpha/2,df}\cdot \sqrt{\frac{s^2}{n_{1}}+\frac{s^2}{n_{2}}} \]

The calculated interval \((-4.044, 0.844)\) suggests that there is a 90% chance that the true difference falls within this range. This interval includes zero, which corresponds to failing to reject the null hypothesis in our test.

Remember, a confidence interval does not tell us the probability of a parameter being inside the interval; rather, it shows the reliability of the estimate. By consistently considering confidence intervals, we can get a clearer picture of our sample data relative to the population.

Pooled Variance

Pooled variance is an essential concept when comparing two populations with potentially different samples sizes. It provides a combined estimate of the variance by pooling data from both samples.

In this exercise, pooled variance is applied since we assume that both populations have the same variance. The formula for pooled variance is: \[ s^2 = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} \]

This formula weighs each sample variance by its degrees of freedom, combining them to form a single, more stable variance estimate. In our case, calculated pooled variance \(s^2=6.514\) aids in computing both the confidence interval and the hypothesis test.

Using pooled variance assumes homogeneity of variance (equal variances) in the two populations. This assumption allows for more efficient inference. Always assess whether this assumption is reasonable in your context. By doing so, you ensure that your statistical conclusions remain valid.