Question

select the correct answer from each drop - down menu. carlos is analyzing the results of an experiment on two groups of mice in which he compared the difference in their weight gain after feeding each group a different type of food for one month. the difference in the mean weight gain is 0.6 grams. the standard deviation of the difference in sample means is 0.305. the 68% confidence interval for the population mean difference is ± dropdown. the 95% confidence interval for the population mean difference is ± dropdown.

Explanation:

Step1: Recall 68-95-99.7 rule

For a normal distribution, 68% of data lies within \( \mu \pm 1\sigma \), 95% within \( \mu \pm 2\sigma \), 99.7% within \( \mu \pm 3\sigma \). Here, mean difference (\( \mu \)) is 0.6, standard deviation of difference in sample means (\( \sigma \)) is 0.305.

Step2: Calculate 68% confidence interval margin

For 68% CI, margin of error is \( 1\times\sigma = 1\times0.305 = 0.305 \).

Step3: Calculate 95% confidence interval margin

For 95% CI, margin of error is \( 2\times\sigma = 2\times0.305 = 0.61 \) (or approximately 0.61, but using 2*0.305 = 0.61).

Answer:

The 68% confidence interval for the population mean difference is \( \pm \boldsymbol{0.305} \). The 95% confidence interval for the population mean difference is \( \pm \boldsymbol{0.61} \) (or \( \pm 2\times0.305 = \pm 0.61 \)).