Examples for 6.2 Confidence Interval AND Hypothesis Test for a Mean Example 1: Dark Chocolate for Good Health Eleven people were given 46 grams (1.6 ounces) of dark chocolate every day for two weeks, and their vascular health was measured before and after the two weeks. Larger numbers indicate greater vascular health, and the mean increase for the participants was 1.3 with a standard deviation of 2.32 . Assume a dotplot shows the data are reasonably symmetric with no extreme values. Find and interpret a $90 \%$ confidence interval for the mean increase in this measure of vascular health after two weeks of eating dark chocolate. Can we be $90 \%$ confident that the mean change for everyone would be positive?

Step 1 ：To find the 90% confidence interval for the mean increase in vascular health, we will use the formula for a confidence interval for a mean when the population standard deviation is unknown and the sample size is small (n < 30). Since the sample size is 11, we will use the t-distribution. The formula for the confidence interval is: \[ \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \] Where: - \( \bar{x} \) is the sample mean - \( t_{\alpha/2} \) is the t-score that cuts off an area of \(\alpha/2\) in the upper tail of the t-distribution with \( n - 1 \) degrees of freedom - \( s \) is the sample standard deviation - \( n \) is the sample size Given: - \( \bar{x} = 1.3 \) - \( s = 2.32 \) - \( n = 11 \) First, we need to find the t-score for a 90% confidence interval with \( n - 1 = 10 \) degrees of freedom. We look up the t-score for a 90% confidence interval and 10 degrees of freedom in a t-distribution table or use a calculator that provides this functionality. The t-score that corresponds to a 90% confidence interval and 10 degrees of freedom is approximately 1.812. Now we can calculate the margin of error (ME): \[ ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.812 \times \frac{2.32}{\sqrt{11}} \approx 1.812 \times 0.699 \approx 1.267 \] The 90% confidence interval is: \[ \bar{x} \pm ME = 1.3 \pm 1.267 \] \[ \text{Lower bound} = 1.3 - 1.267 = 0.033 \] \[ \text{Upper bound} = 1.3 + 1.267 = 2.567 \] So the 90% confidence interval for the mean increase in vascular health after two weeks of eating dark chocolate is (0.033, 2.567). Interpretation: We are 90% confident that the true mean increase in vascular health for the entire population, after two weeks of eating dark chocolate, lies between 0.033 and 2.567. Since the entire interval is above zero, we can be 90% confident that the mean change for everyone would be positive. Regarding the hypothesis test, if we were to conduct one, we would set up a null hypothesis that there is no increase in vascular health (mean increase = 0) and an alternative hypothesis that there is an increase (mean increase > 0). Given that our confidence interval does not include zero and is entirely above zero, we would reject the null hypothesis at the 90% confidence level and conclude that there is evidence to suggest that eating dark chocolate does lead to a positive mean increase in vascular health. However, to make a formal conclusion, we would need to perform the hypothesis test with an appropriate significance level and calculate the p-value.