Home / AP® Exam / AP® Statistics / AP Statistics 4.5 Carrying Out a Test for a Population Mean or Population Mean Difference- Exam Style Questions – MCQs

AP Statistics 4.5 Carrying Out a Test for a Population Mean or Population Mean Difference- Exam Style Questions - MCQs - New Syllabus

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Question 

At a local arcade, one of the games involves spinning a wheel with twelve equal sections in which the wheel lists the number of tickets that the player wins. One section has “500,” two sections have “100,” three sections have “50,” four sections have “25,” and two sections have “5.” Assume each section is equally likely. If each play costs 4 tokens, how many tokens would a player expect to use to land on the 500 tickets?

(A) 4
(B) 12
(C) 48
(D) 52
(E) 500

▶️ Answer/Explanation

The probability of landing on the 500-ticket section is:
\( p=\frac{1}{12} \)

The number of plays needed to get the first 500-ticket result follows a geometric distribution. The expected number of plays for a geometric random variable is:
\( E(X)=\frac{1}{p} \)
\( =\frac{1}{1/12} \)
\( =12 \text{ plays} \)

Each play costs 4 tokens, so the expected number of tokens used is:
\( 12 \times 4 = 48 \)

Therefore, a player would expect to spend 48 tokens before landing on the 500-ticket section.

✅ Answer: (C)

Question 

An independent research firm conducted a study of 100 randomly selected children who were participating in a program advertised to improve mathematics skills. The results showed no statistically significant improvement in mathematics skills, using \(\alpha = 0.05\). The program sponsors complained that the study had insufficient statistical power. Assuming that the program is effective, which of the following would be an appropriate method for increasing power in this context.

(A) Use a two-sided test instead of a one-sided test.
(B) Use a one-sided test instead of a two-sided test.
(C) Use \(\alpha = 0.01\) instead of \(\alpha = 0.05\).
(D) Decrease the sample size to 50 children.
(E) Decrease the sample size to 200 children.

▶️ Answer/Explanation

Statistical power is the probability of correctly rejecting a false null hypothesis. Since the sponsors believe the program is effective, increasing power would make it more likely that the study detects a real improvement in mathematics skills.

One of the most effective ways to increase power is to increase the sample size. A larger sample size reduces the standard error, making it easier to distinguish a true effect from random variation.

The standard error decreases as sample size increases:
\( SE \propto \frac{1}{\sqrt{n}} \)
Increasing the sample size from
\( n=100 \) to \( n=200 \)
results in a smaller standard error and greater statistical power.

Using a smaller significance level such as \( \alpha=0.01 \) would reduce power, and decreasing the sample size would also reduce power.

Therefore, the appropriate method for increasing power is to increase the sample size to 200 children.

✅ Answer: (E)

Question

A medical researcher completed a matched-pairs study comparing an omega-\(3\) supplement to a placebo in the treatment of irritability. For each of \(19\) patients, the researcher recorded the difference in irritability score, \(d=\text{placebo score}-\text{omega-}3\text{ score}\).
The summary statistics for the differences are shown below.
Which of the following gives the correct test statistic and degrees of freedom for testing \(H_0:\mu_d=0\) against \(H_a:\mu_d>0\)?
(A) \(t=\dfrac{1.789-0}{2.485}\approx0.720\), with \(df=19\)
(B) \(t=\dfrac{1.789-0}{2.485/\sqrt{19}}\approx3.138\), with \(df=18\)
(C) \(z=\dfrac{1.789-0}{2.485/\sqrt{19}}\approx3.138\), with \(df=18\)
(D) \(t=\dfrac{5.421-3.632}{2.987/\sqrt{19}}\approx2.612\), with \(df=18\)
▶️ Answer/Explanation

For a matched-pairs \(t\)-test, the test statistic is calculated using the sample mean difference, the sample standard deviation of the differences, and the number of pairs.
\(t=\dfrac{\bar{x}_d-\mu_0}{s_d/\sqrt{n}}\)
Substituting the values:
\(t=\dfrac{1.789-0}{2.485/\sqrt{19}}\)
\(t\approx3.138\)

Since there are \(n=19\) paired observations, the degrees of freedom are:
\(df=n-1=19-1=18\)

✅ Answer: (B)

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