Question

An experiment was conducted to compare two diets \(\mathrm{A}\) and \(\mathrm{B}\) designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet \(\mathrm{A}\) and the other on diet \(\mathrm{B},\) and their weight losses were recorded over a 30 -day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a \(95 \%\) confidence interval for the difference in mean weight loss for the two diets. Interpret your confidence interval. $$\begin{array}{ll}\text { Diet } A & \text { Diet B } \\\\\hline \bar{x}_{A}=21.3 & \bar{x}_{B}=13.4 \\\s_{A}=2.6 & s_{B}=1.9\end{array}$$

Short answer

Answer: The 95% confidence interval for the difference in mean weight loss for diets A and B is (6.5, 9.3) pounds.

Step-by-step solution

Identify given statistics

We are provided with the following statistics: - \(\bar{x}_{A} = 21.3\), \(\bar{x}_{B} = 13.4\) (sample means) - \(s_{A} = 2.6\), \(s_{B} = 1.9\) (sample standard deviations) - \(n_A = n_B = 30\) (sample sizes) Our goal is to find the 95% confidence interval for the difference in mean weight loss, \(μ_A - μ_B\).

Find the standard error

The standard error for the difference in means is given by: \(SE_{\bar{X_A} - \bar{X_B}} = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}\) Now, plug in the given values: \(SE_{\bar{X_A} - \bar{X_B}} = \sqrt{\frac{2.6^2}{30} + \frac{1.9^2}{30}} \approx 0.714\)

Find the critical value

For a 95% confidence interval, we will use a critical value from a standard normal distribution (Z-distribution) table, since the sample size is large enough: \(z_{\frac{1-0.95}{2}} = z_{0.025} = 1.96\)

Calculate the margin of error

The margin of error is the product of the critical value and the standard error: \(MOE = z_{0.025} \cdot SE_{\bar{X_A} - \bar{X_B}} = 1.96 \cdot 0.714 \approx 1.40\)

Find the confidence interval

Finally, we can find the 95% confidence interval for the difference in mean weight loss for the two diets: \(CI_{95\%} = (\bar{x}_{A} - \bar{x}_{B}) \pm MOE\) \(CI_{95\%} = (21.3 - 13.4) \pm 1.40\) \(CI_{95\%} = 7.9 \pm 1.40\) So, the confidence interval is \((6.5, 9.3)\). In conclusion, we are 95% confident that the difference in mean weight loss for diet A and diet B lies between 6.5 and 9.3 pounds, with diet A being more effective for weight reduction.

Key concepts

Difference in Means

When contrasting two groups, such as those on diet A and diet B, the difference in means is a crucial statistic. In this context, it refers to the average weight loss difference between the two groups. Essentially, we subtract the mean weight loss of diet B from diet A to determine how much more effective diet A might be than diet B. If we look at our experiment, this calculation is represented by \( \bar{x}_A - \bar{x}_B = 21.3 - 13.4 = 7.9 \). This means that diet A's participants, on average, lost 7.9 pounds more than those on diet B. Calculating the difference in means helps researchers understand the magnitude of increase or decrease between treatments or interventions. This difference serves as an anchor point for creating confidence intervals.

Standard Error

The standard error is a measure of the variability or dispersion of the sample mean's distribution. For our study involving weight loss on diets A and B, the standard error helps us quantify the precision of our sample mean difference. The formula to calculate the standard error for the difference in means is:

• \( SE_{\bar{X}_A - \bar{X}_B} = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}} \)

In our case, substituting our values gives \( SE_{\bar{X}_A - \bar{X}_B} \approx 0.714 \). This means the expected variation or uncertainty around our calculated mean difference is approximately 0.714 pounds, reflecting both sample sizes and standard deviations of the groups. A smaller standard error indicates a more precise estimate of the population parameter, reinforcing confidence in our findings.

Z-distribution

The Z-distribution, also known as the standard normal distribution, is symmetrical and centered at zero. It is extensively used in statistics for hypothesis testing, particularly when the sample size is large. In our experiment involving diets, the Z-distribution helps us determine critical values that define the margin of error for our confidence interval.

• For a confidence level of 95%, we look for the Z-value representing the cutoff in the table, typically found at \( z_{0.025} = 1.96 \).

This Z-value corresponds to the middle 95% of the distribution, with 2.5% in each tail, establishing the limits for our hypothesis test. Using the Z-distribution is convenient as it assumes sample means tend to distribute normally if the sample size is reasonably large, due to the Central Limit Theorem.

Margin of Error

The margin of error expresses the range within which we expect the true difference in means to fall. It's calculated by multiplying the critical value from the Z-distribution by the standard error. In our example involving diets:

• \( MOE = 1.96 \times 0.714 \approx 1.40 \)

This result means we can be confident that the true mean difference is within ±1.40 pounds of our observed difference. Hence, our confidence interval ranges from 6.5 to 9.3 pounds. A smaller margin of error signifies greater precision of our estimates, hence a more reliable conclusion about the comparative effectiveness of the studied diets. Always remember, a confidence interval that does not include zero reinforces the presence of a true difference.