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AP Statistics – Inference Practice

Confidence intervals, significance tests, chi-square, regression inference.

12 questions  ·  74 marks  ·  answers included  ·  AP Statistics
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This worksheet covers:

  • Confidence intervals and significance tests for proportions and means (one-sample, two-sample and paired)
  • Chi-square tests
  • Inference for the slope of a regression line

A graphing calculator is allowed; state conditions, name the procedure, and interpret conclusions in context. Allow about 90 minutes.

Questions

Section A – Skills

Give calculated values to 3 decimal places.
1

In a random sample of $400$ adults from a city, $120$ said they cycle to work at least once a week.(a) Verify that the conditions for a one-sample $z$-interval for a proportion are met.(b) Construct and interpret a $95$ percent confidence interval for the proportion of all adults in the city who cycle to work at least once a week.

Show answer
(a) Random sample; $n\hat{p} = 120 \geq 10$ and $n(1-\hat{p}) = 280 \geq 10$; population $> 4000$ (10 percent condition)(b) $0.3 \pm 1.96\sqrt{\dfrac{0.3(0.7)}{400}} = 0.3 \pm 0.045$, i.e. $(0.255,\ 0.345)$. We are 95 percent confident the true proportion is between $0.255$ and $0.345$.
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2

A random sample of $25$ packets of coffee has mean mass $52.4$ g and standard deviation $6.8$ g. A dotplot of the sample shows no strong skew or outliers.(a) Explain why a $t$-procedure is appropriate.(b) Construct a $90$ percent confidence interval for the mean mass of all packets.

Show answer
(a) Population SD unknown; sample is random; $n < 30$ but the dotplot shows no strong skew or outliers, so Normality is plausible(b) $t^* = 1.711$ (df $= 24$): $52.4 \pm 1.711\dfrac{6.8}{\sqrt{25}} = 52.4 \pm 2.33$, i.e. $(50.1,\ 54.7)$ g
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3

A polling organisation wants to estimate the proportion of voters who support a proposal to within $\pm 0.03$ with $95$ percent confidence. No prior estimate is available.Find the minimum sample size required.

Show answer
$n = \left(\dfrac{1.96}{0.03}\right)^2 (0.5)(0.5) = 1067.1$, so $n = 1068$
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4

A company claims that $60$ percent of its customers renew their subscription. In a random sample of $400$ customers, $210$ renewed.(a) State the hypotheses for a test of the company’s claim.(b) Calculate the test statistic and $P$-value.(c) State your conclusion at $\alpha = 0.05$.

Show answer
(a) $H_0: p = 0.6$, $H_a: p eq 0.6$(b) $z = \dfrac{0.525 – 0.6}{\sqrt{0.6(0.4)/400}} = -3.06$, $P \approx 0.0022$(c) $P < 0.05$: reject $H_0$; convincing evidence the renewal rate differs from $60$ percent
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5

A machine is set to fill bottles with $500$ mL. A random sample of $36$ bottles has mean $492$ mL and standard deviation $18$ mL.(a) Calculate the $t$ statistic for testing $H_0: \mu = 500$ against $H_a: \mu < 500$.(b) Find the $P$-value and state a conclusion at the $1$ percent significance level.

Show answer
(a) $t = \dfrac{492 – 500}{18/\sqrt{36}} = -2.667$, df $= 35$(b) $P \approx 0.0057 < 0.01$: reject $H_0$; convincing evidence the mean fill is below $500$ mL
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Section B – Exam-style

Full four-step procedure expected where a test or interval is requested.
6

In a randomised experiment, $150$ patients received a new drug and $150$ received a placebo. $45$ of the drug group and $30$ of the placebo group reported improvement.Carry out a test at $\alpha = 0.05$ to determine whether the drug increases the proportion of patients who improve.

Show answer
$H_0: p_1 = p_2$, $H_a: p_1 > p_2$; random assignment, all counts $\geq 10$$\hat{p}_1 = 0.3$, $\hat{p}_2 = 0.2$, pooled $\hat{p} = 0.25$$z = \dfrac{0.1}{\sqrt{0.25(0.75)(2/150)}} = 2.00$, $P \approx 0.023$Reject $H_0$: convincing evidence the drug increases the improvement rate
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7

Twenty students took a reaction-time test before and after drinking a caffeinated drink. For each student the difference (before minus after, in hundredths of a second) was computed. The mean difference was $2.3$ with standard deviation $4.1$; a plot of the differences was roughly symmetric.(a) Explain why a paired $t$-test is appropriate rather than a two-sample $t$-test.(b) Test whether caffeine reduces reaction time, using $\alpha = 0.05$.

Show answer
(a) Each before/after pair comes from the same student, so the observations are not independent samples(b) $H_0: \mu_d = 0$, $H_a: \mu_d > 0$; $t = \dfrac{2.3}{4.1/\sqrt{20}} = 2.51$, df $= 19$, $P \approx 0.011$. Reject $H_0$: convincing evidence caffeine reduces reaction time
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8

A six-sided die is rolled $120$ times. The observed frequencies of $1, 2, 3, 4, 5, 6$ are $15, 22, 18, 25, 20, 20$.(a) State the hypotheses for a chi-square goodness-of-fit test.(b) Calculate the chi-square statistic and its degrees of freedom.(c) Find the $P$-value and state a conclusion at $\alpha = 0.10$.

Show answer
(a) $H_0$: the die is fair (each face has probability $\frac{1}{6}$); $H_a$: at least one probability differs(b) Expected $= 20$ each; $\chi^2 = \dfrac{25 + 4 + 4 + 25 + 0 + 0}{20} = 2.9$, df $= 5$(c) $P \approx 0.715$; fail to reject $H_0$; no convincing evidence the die is unfair
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9

A random sample of $120$ shoppers was classified by gender and preferred store layout (A, B or C). Males: A $30$, B $20$, C $10$. Females: A $20$, B $30$, C $10$.(a) Find the expected count for males preferring layout A.(b) Calculate the chi-square statistic and degrees of freedom for a test of independence.(c) State the $P$-value and conclusion at $\alpha = 0.05$.

Show answer
(a) $\dfrac{60 \times 50}{120} = 25$(b) Expected counts $25, 25, 10$ in each row; $\chi^2 = 4 \times \dfrac{25}{25} = 4.0$, df $= 2$(c) $P \approx 0.135$; fail to reject $H_0$; no convincing evidence of an association between gender and layout preference
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10

Computer output for a regression of exam score on hours studied ($n = 20$ students) gives slope $b = 2.45$ with standard error $0.62$. Conditions for inference are satisfied.(a) Construct a $95$ percent confidence interval for the population slope $\beta$.(b) Test $H_0: \beta = 0$ against $H_a: \beta eq 0$ and interpret the result.

Show answer
(a) $t^* = 2.101$ (df $= 18$): $2.45 \pm 2.101(0.62) = 2.45 \pm 1.30$, i.e. $(1.15,\ 3.75)$(b) $t = \dfrac{2.45}{0.62} = 3.95$, $P \approx 0.001$; reject $H_0$; convincing evidence of a linear relationship; each extra hour is associated with about $2.45$ more points
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Section C – Free-response

Investigative tasks.
11

Two teaching methods were compared. Method A: $n = 30$, $\bar{x} = 78.2$, $s = 8.1$. Method B: $n = 28$, $\bar{x} = 74.5$, $s = 9.4$. Students were randomly assigned to methods.(a) Calculate the standard error of $\bar{x}_A – \bar{x}_B$.(b) Carry out a two-sample $t$-test (using the conservative df $= 27$ or technology) to determine whether the methods differ in mean score, at $\alpha = 0.05$.(c) Describe a Type I error in this context.

Show answer
(a) $\sqrt{\dfrac{8.1^2}{30} + \dfrac{9.4^2}{28}} = 2.31$(b) $H_0: \mu_A = \mu_B$, $H_a: \mu_A eq \mu_B$; $t = \dfrac{3.7}{2.31} = 1.60$; $P \approx 0.115$ (df $\approx 53.5$) or $P \approx 0.12$ (df $= 27$); fail to reject $H_0$; no convincing evidence of a difference(c) Concluding the methods differ in mean score when in fact they do not
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12

Historically $20$ percent of applicants to a programme are accepted. A new admissions officer accepts $16$ of a random sample of $50$ applicants from this year.(a) Check whether a one-proportion $z$-test is appropriate.(b) Test whether the acceptance rate has increased, at $\alpha = 0.05$.(c) Explain what is meant by the power of this test, and give one way to increase it.

Show answer
(a) Random sample; $np_0 = 10$ and $n(1 – p_0) = 40$, both $\geq 10$; population $> 500$(b) $H_0: p = 0.2$, $H_a: p > 0.2$; $\hat{p} = 0.32$; $z = \dfrac{0.32 – 0.2}{\sqrt{0.2(0.8)/50}} = 2.12$, $P \approx 0.017$; reject $H_0$; convincing evidence the rate has increased(c) Power is the probability of correctly rejecting $H_0$ when a specific alternative is true; increase $n$ or $\alpha$
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