Home / AP® Exam / AP® Statistics / 7.3 Justifying a Claim About a Population Mean Based on a Confidence Interval- FRQs

AP Statistics 7.3 Justifying a Claim About a Population Mean Based on a Confidence Interval- FRQs - Exam Style Questions

AP Statistics -FRQs and Answers - All Topics

Question

According to a \(2017\) national survey in Country B, the mean number of bedrooms in newly built houses was \(2.9\). Rodney, a researcher, believes the mean number of bedrooms in newly built houses in the country was different in \(2024\) than it was in \(2017\). To investigate his belief, he took a large random sample of newly built houses in Country B in \(2024\) and recorded the number of bedrooms in each house. The distribution of the number of bedrooms for the sampled houses is summarized in the table.
Number of Bedrooms\(1\)\(2\)\(3\)\(4\)\(5\)\(6\)
Proportion of Houses\(0.12\)\(0.22\)\(0.28\)\(0.22\)\(0.14\)\(0.02\)
A.
i. A house from the sample will be selected at random. What is the probability that the house had fewer than \(3\) bedrooms? Show your work.
ii. What is the mean number of bedrooms for the sample of newly built houses in \(2024\)? Show your work.
B. Rodney will use a one-sample \(t\)-test for a population mean to test his belief.
i. In the context of Rodney’s investigation, state the hypotheses for the test.
ii. Explain, in context, what a Type I error would be for Rodney’s hypothesis test.
C. A different researcher, Keisha, suggests using a confidence interval to investigate whether the mean number of bedrooms in newly built houses in \(2024\) in Country B was different from \(2.9\).
Assume the conditions for inference have been met. Using Rodney’s data, Keisha calculated a one-sample \(97\) percent confidence interval to estimate the population mean as \((3.01, 3.19)\). Based on the confidence interval, what conclusion can be made for Rodney’s hypothesis test in part B at \(\alpha = 0.03\)? Justify your answer.

Most-appropriate topic codes (CED):

• TOPIC 4.7: Introduction to Random Variables and Probability Distributions — part (A)
• TOPIC 7.4: Setting Up a Test for a Population Mean — part (B i)
• TOPIC 7.8: Potential Errors When Performing Tests — part (B ii)
• TOPIC 7.3: Justifying a Claim About a Population Mean Based on a Confidence Interval — part (C)
▶️ Answer/Explanation
Detailed solution

A i.
Let \(X\) be the number of bedrooms in a randomly selected house.
\(P(X < 3) = P(X = 1) + P(X = 2) = 0.12 + 0.22 = 0.34\).
\(\boxed{0.34}\)

A ii.
The mean is the expected value of \(X\):
\[E(X) = \sum x \cdot P(X = x) = 1(0.12) + 2(0.22) + 3(0.28) + 4(0.22) + 5(0.14) + 6(0.02)\]
\[E(X) = 0.12 + 0.44 + 0.84 + 0.88 + 0.70 + 0.12 = 3.10\]
\(\boxed{3.10}\)

B i.
Let \(\mu\) be the population mean number of bedrooms in newly built houses in Country B in \(2024\).
\(H_0: \mu = 2.9\)
\(H_a: \mu \neq 2.9\)

B ii.
A Type I error would occur if Rodney concludes that the population mean number of bedrooms in newly built houses in \(2024\) is different from \(2.9\) when, in reality, it is \(2.9\).

C.
The \(97\%\) confidence interval is \((3.01, 3.19)\). Since the hypothesized value \(2.9\) is not contained in this interval, we reject \(H_0: \mu = 2.9\) at the \(\alpha = 0.03\) significance level.
There is convincing statistical evidence that the population mean number of bedrooms in newly built houses in Country B in \(2024\) is different from \(2.9\).

More AP Statistics FRQs Questions..
Scroll to Top