AP Statistics 4.6 Sampling Distributions for the Difference Between Two Sample Means- Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Since the population standard deviation is unknown and the sample size is small (\(n=18\)), a t-interval should be used.
The general form of a confidence interval for a population mean is:
\( \bar{x} \pm t^*\left(\frac{s}{\sqrt{n}}\right) \)
For a 95% confidence interval with
\( n = 18 \),
degrees of freedom:
\( df = 18 – 1 = 17 \)
and
\( t^* \approx 2.11 \).
Substituting the given values:
\( 64.6 \pm 2.11\left(\frac{11.7}{\sqrt{18}}\right) \)
Therefore, the correct confidence interval expression is choice (C).
✅ Answer: (C)
▶️ Answer/Explanation
One important condition for a two-sample t-test is that the sample sizes are large enough, or that the data in each group come from populations that are approximately Normal. Large sample sizes help satisfy the conditions needed for valid inference.
The other choices discuss characteristics such as age distributions, dietary preferences, or correlations that are not required assumptions for conducting a two-sample t-test.
Before carrying out the test, statisticians typically check for independence and whether the sample sizes are sufficiently large (or whether the Normal condition is reasonable).
✅ Answer: (B)
▶️ Answer/Explanation
The response variable is number of hours per week consuming news media, which is a quantitative variable. The explanatory variable is whether or not an adult has children, which creates two groups: adults with children and adults without children.
The research question asks whether the mean number of hours spent consuming news media differs between these two groups. When comparing the means of a quantitative variable across two independent groups, the appropriate procedure is a two-sample t-test for means.
The other procedures are designed for proportions or for relationships between categorical variables, which do not match the variable types in this problem.
✅ Answer: (D)