AP Statistics 4.2 Constructing a Confidence Interval for a Population Mean or Population Mean Difference- Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Since the population standard deviation is unknown and the sample size is small (\(n=16\)), a t-interval should be used.
Degrees of freedom:
\(df=n-1=15\)
For a 95% confidence interval with \(df=15\):
\(t^* \approx 2.131\)
The standard error is:
\(\frac{s}{\sqrt{n}}=\frac{7.8}{\sqrt{16}}=\frac{7.8}{4}\)
Therefore, the confidence interval is:
\(36.4 \pm 2.131\left(\frac{7.8}{4}\right)\) which corresponds to choice (A).
✅ Answer: (A)
▶️ 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
Let \(p\) represent the true proportion of all critics who would give the Season 2 finale a favorable rating.
The network will renew the show only if more than 60% of critics approve. Therefore, the claim the producers want to support is:
\( p > 0.60 \)
In a significance test, the null hypothesis contains the equality:
\( H_0: p = 0.60 \)
The alternative hypothesis reflects the claim of interest:
\( H_a: p > 0.60 \)
This is a right-tailed one-proportion z-test because the question asks whether there is evidence that the approval rate exceeds 60%.
✅ Answer: (D)