Question

Suppose that a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\) is calculated to be (1.4,6.7) . If we test \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{A}: \mu_{1}-\mu_{2} \neq 0\) using \(\alpha=0.05\) will we reject \(H_{0} ?\) Why or why not?

Short answer

We will reject the null hypothesis \(H_0\) since 0 is not within the 95% confidence interval (1.4, 6.7).

Step-by-step solution

Identify the Confidence Interval and Hypothesis

The provided confidence interval is (1.4, 6.7) for the difference in means \(\mu_1 - \mu_2\). The null hypothesis \(H_0\) states that \(\mu_1 - \mu_2 = 0\), and it will be tested against the alternative hypothesis \(H_A\) that claims \(\mu_1 - \mu_2 eq 0\), with a significance level of \(\alpha = 0.05\).

Compare Hypothesized Value with Confidence Interval

Since the null hypothesis proposes no difference between the means, we compare the hypothesized difference of 0 to the confidence interval. If 0 lies outside of the interval (1.4, 6.7), it suggests that the null hypothesis is unlikely.

Make a Decision Regarding the Null Hypothesis

The value 0 is not included in the interval (1.4, 6.7), therefore we reject the null hypothesis \(H_0\) because the confidence interval for \(\mu_1 - \mu_2\) does not contain the hypothesized value of 0.

Key concepts

Confidence Interval

When we talk about a confidence interval, we're discussing a range of values, estimated from the data, that is likely to contain the true population parameter, such as the population mean, at a given confidence level. For example, a 95% confidence interval for the difference between two population means, \(\mu_1 - \mu_2\), suggests that if we took many samples and built an interval from each sample, about 95% of these intervals would contain the true difference \(\mu_1 - \mu_2\). In our exercise, the interval is (1.4, 6.7), which means we can be 95% confident that the true difference in means lies within this range. Since 0 is not within this interval, we can infer that the true difference is likely non-zero, pointing towards a statistically significant difference between the two means.

Hypothesis Testing

Hypothesis testing is a formal procedure used by statisticians to either accept or reject statistical hypotheses. The null hypothesis, denoted as \(H_0\), is the default assumption that there is no effect or no difference. Alternatively, the alternative hypothesis, \(H_A\), is what researchers want to prove — that there is an effect or a difference. In the context of the given exercise, we have \(H_0: \mu_1 - \mu_2 = 0\) implying no difference, and \(H_A: \mu_1 - \mu_2 eq 0\) implying a difference exists. We use the calculated confidence interval to assess the validity of \(H_0\). Since the confidence interval does not contain the value 0, we reject \(H_0\) and accept \(H_A\), leading us to conclude that there is evidence of a difference between the two means.

Difference in Means

The difference in means is a statistical measure used frequently to compare two groups. When we calculate \(\mu_1 - \mu_2\), we are looking at how much the average outcome of one group differs from another. This is particularly useful in experiments and observational studies to determine if there is a significant effect of a treatment or a significant association between variables. In our exercise, the confidence interval for the difference in means, (1.4, 6.7), does not include 0, indicating that \(\mu_1\) is likely greater than \(\mu_2\), and hence, there is a statistically significant difference between the two group means.

Significance Level

The significance level, denoted as \(\alpha\), is the threshold at which we decide whether or not to reject the null hypothesis. It represents the probability of making a Type I error, which occurs when we incorrectly reject a true null hypothesis. Common choices for \(\alpha\) include 0.05, 0.01, and 0.10. In hypothesis testing, if the p-value is less than \(\alpha\), we reject the null hypothesis. Given an \(\alpha\) of 0.05 in our exercise and the fact that our confidence interval does not contain the hypothesized value (meaning our p-value is less than 0.05), we therefore reject the null hypothesis, suggesting that the difference in means is significant at the 5% level.