## MATH1015 Quizzes

Quiz 10: Paired and two-sample tests for means
Question 1 Questions
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
To test whether the diet is successful, state the appropriate null and alternative hypotheses in terms of ${p}_{+}$ (the probability that the diet increases weight). Exactly one option must be correct)
 a) ${H}_{0}:{p}_{+}=0;\phantom{\rule{1em}{0ex}}{H}_{1}:{p}_{+}<0$. b) ${H}_{0}:{p}_{+}=0.5;\phantom{\rule{1em}{0ex}}{H}_{1}:{p}_{+}<0.5$. c) ${H}_{0}:{p}_{+}=0.5;\phantom{\rule{1em}{0ex}}{H}_{1}:{p}_{+}\ne 0.5$. d) ${H}_{0}:{p}_{+}=0.5;\phantom{\rule{1em}{0ex}}{H}_{1}:{p}_{+}>0.5$.

Choice (a) is incorrect
A zero change in weight is not the same as ${p}_{+}=0$.
Choice (b) is correct!
We are looking for evidence of a decrease in weight on average, so we expect the proportion with a positive weight change to be below 0.5.
Choice (c) is incorrect
This is not a two-sided test. What is the anticipated alternative?
Choice (d) is incorrect
What is the anticipated alternative in terms of ${p}_{+}$?
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
Is there evidence that the diet is successful? Provide a P-value and state your findings. Exactly one option must be correct)
 a) P-value = 0.1133. There is insufficient evidence to claim that the diet is successful. b) P-value = 0.9673. The data do not provide evidence that the diet is successful. c) P-value = 0.0302. There is very strong evidence that the diet is successful. d) P-value = 0.1133. The probability that the diet is successful is 0.1133.

Choice (a) is correct!
Write $\tau$ to represent the number of weight increases. Under ${H}_{0}$, $\tau \sim \mathsc{ℬ}\left(11,0.5\right)$. Since this is a lower tail test, and since the observed value of the test statistic is ${\tau }_{obs}=3$, the P-value is $P\left(\tau \le 3\right)=0.1133$.
Choice (b) is incorrect
The P-value calculates the probability of a result at least as extreme as the observed result of 3 positive weight gains.
Choice (c) is incorrect
The P-value calculates the probability of a result at least as extreme as the observed result of 3 positive weight gains.
Choice (d) is incorrect
Revise the interpretation of a P-value.
Questions 1 - 3 use the following information.
Eleven patients were weighed before and after a diet which was designed to help patients reduce weight. Eight of these patients did lose weight but three patients had a weight increase.
The observed sample mean (in kg) and sample variance (in kg${}^{2}$) of the eleven changes in weight (after - before) were $\stackrel{̄}{x}=-0.6$ and ${s}^{2}=1.44$. Writing ${\mu }_{d}$ to represent the mean change in weight (after - before), provide a P-value for testing ${H}_{0}:{\mu }_{d}=0$ vs ${H}_{1}:{\mu }_{d}<0$ assuming that the weight changes follow a normal distribution. (Round ${\tau }_{obs}$ to 2dp.) Exactly one option must be correct)
 a) P-value = $P\left(Z\le -1.66\right)\approx 0.0485$ b) P-value= $P\left({t}_{11}<-1.66\right)>0.05$. c) P-value = $P\left({t}_{10}<-1.38\right)>0.05$ d) P-value = $P\left({t}_{10}<-1.66\right)>0.05$

Choice (a) is incorrect
The population standard deviation of the differences is unknown, so you should not use the Z-test.
Choice (b) is incorrect
You have used the wrong $t$-distribution.
Choice (c) is incorrect
Have you used ${s}^{2}$ by mistake, instead of $s$ in the calculation of the observed test statistic?
Choice (d) is correct!
${\tau }_{obs}=\frac{-0.6}{1.2∕\sqrt{11}}=-1.66$, and P-value = $P\left({t}_{10}\le -1.66\right)$ which is in the interval (0.05, 0.1).
Twelve athletes were each given two different exercise routines, and their heartbeats were monitored closely. The aim of the study was to test for a difference in the effect of the routines on heartbeat rate. The twelve differences were (in appropriate units):
${d}_{i}$: 1.5 2.3  4.7  6.1  -1.2  2.6  1.5  -0.4  -2.7  1.9  1.1  -1.5.
(Elementary calculations give: ${\sum }_{i}{d}_{i}=15.9$ and ${\sum }_{i}{d}_{i}^{2}=91.81$.) Assuming the differences are normally distributed, provide bounds for the P-value from the appropriate $t$-test. Exactly one option must be correct)
 a) (0.025, 0.05) b) (0.05, 0.10) c) (0.10, 0.20) d) (0.01, 0.025)

Choice (a) is incorrect
This is a two-sided test.
Choice (b) is correct!
The sample mean difference is 1.325 and the sample standard deviation of the differences is 2.536, so ${\tau }_{obs}=\frac{1.325}{2.536∕\sqrt{12}}=1.809$. The test is two-sided, so the P-value = $P\left(|{t}_{11}|>1.809\right)=2P\left({t}_{11}>1.809\right)$, which is in the interval (0.05, 0.10).
Choice (c) is incorrect
Check your calculation of the sample mean and standard deviation.
Choice (d) is incorrect
Check your calculation of the sample mean and standard deviation.
A panel of psychiatrists is asked to compare the condition of 65 patients before and after intermittent psychotherapy, which is expected to improve the condition in general. 40 patients showed an improvement, 24 were worse and one was considered unchanged. Use the sign test with the normal approximation to find the P-value. Exactly one option must be correct)
 a) $P\left(X\ge 40\right)=0$ b) $P\left(Z\ge 40\right)=0$ c) $P\left(Z\ge 2\right)=0.0228$ d) $P\left(Z>1.875\right)\approx 0.03$

Choice (a) is incorrect
You need the approximating normal variable to find $P\left(X\ge 40\right)$.
Choice (b) is incorrect
You need to use the approximating normal variable and standardise it.
Choice (c) is incorrect
Perhaps you have forgotten the continuity correction.
Choice (d) is correct!
Ignore the one unchanged patient, and let $X$ represent the number improved out of 64. Under the hypothesis of no improvement, $X\sim \mathsc{ℬ}\left(64,0.5\right)$ and the P-value is $P\left(X\ge 40\right)$, since this is an upper-tail test. Using a normal approximation with correction for continuity, P-value $\approx P\left(Y\ge 39.5\right)$, where $Y\sim \mathsc{N}\left(32,16\right)$ is the approximating normal variable. Thus, P-value $\approx P\left(Z\ge \frac{39.5-32}{\sqrt{16}}\right)=P\left(Z\ge 1.875\right)\approx 0.03$.
A test of ${H}_{0}:{\mu }_{x}={\mu }_{y}$ vs ${H}_{1}:{\mu }_{x}\ne {\mu }_{y}$ based on independent samples of sizes ${n}_{x}=12$ and ${n}_{y}=10$, yields a two-sample $t$-statistic with observed value, ${\tau }_{obs}=1.95$. The P-value is in the interval: Exactly one option must be correct)
 a) (0, 0.01) b) (0.01,0.025) c) (0.025,0.05) d) (0.05,0.10)

Choice (a) is incorrect
Check the tables again.
Choice (b) is incorrect
Check the tables again.
Choice (c) is incorrect
Recall that this is a two-tail test.
Choice (d) is correct!
P-value = $P\left(|{t}_{20}|\ge 1.95\right)=2P\left({t}_{20}\ge 1.95\right)$. Since $P\left({t}_{20}\ge 1.95\right)$ is in the interval (0.025, 0.05), the P-value is in the interval (0.05, 0.1).
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: $\stackrel{̄}{x}=129.44$, $ȳ=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.
The pooled estimate of the standard deviation, ${s}_{p}$, in the two-sample test statistic is: Exactly one option must be correct)
 a) 10.3 b) 105.72 c) 106.01 d) 10.28

Choice (a) is correct!
${s}_{p}=\sqrt{\frac{9×9.1{5}^{2}+13×11.0{2}^{2}}{10+14-2}}=10.30$.
Choice (b) is incorrect
Use the formula for ${s}_{p}^{2}$ and then find its square root.
Choice (c) is incorrect
This is the pooled estimate of the variance, not of the standard deviation.
Choice (d) is incorrect
Check the formula for ${s}_{p}^{2}$ and then square root.
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: $\stackrel{̄}{x}=129.44$, $ȳ=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.
To test for equal means, the appropriate $t$-distribution is Exactly one option must be correct)
 a) ${t}_{24}$ b) ${t}_{22}$ c) ${t}_{10}$ - ${t}_{14}$. d) ${t}_{23}$

Choice (a) is incorrect
You do not simply add the sample sizes 10+14 to give the df of the $t$-distribution.
Choice (b) is correct!
The appropriate $t$-variable has df = $10+14-2=22$.
Choice (c) is incorrect
Try again.
Choice (d) is incorrect
With two independent samples of sizes $m$ and $n$, df = $m+n-2$.
Questions 7 - 9 use the following information.
In an attempt to compare the durability of two different materials (X and Y), 10 pieces of type X and 14 pieces of type Y were used. From these samples we calculate the statistics: $\stackrel{̄}{x}=129.44$, $ȳ=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.
Assuming the two populations sampled are normal with equal variance, find the P-value. Exactly one option must be correct)
 a) (0.05, 0.10) b) (0.1, 0.2) c) (0.01, 0.025) d) (0.02, 0.05).

Choice (a) is incorrect
This is a two-sided test.
Choice (b) is correct!
${s}_{p}=\sqrt{\frac{9×9.1{5}^{2}+13×11.0{2}^{2}}{10+14-2}}=10.30$, so ${\tau }_{obs}=\frac{129.44-122.65}{10.30\sqrt{\frac{1}{10}+\frac{1}{14}}}=1.52$.
P-value = $P\left(|{t}_{22}|\ge 1.52\right)=2P\left({t}_{22}\ge 1.52\right)$, with bounds (0.1, 0.2), since $P\left({t}_{22}\ge 1.52\right)$ is in the range (0.05, 0.10).
Choice (c) is incorrect
Try again. Check that ${s}_{p}=10.30$ and that ${\tau }_{obs}=1.52$. Use a two-sided test.
Choice (d) is incorrect
Try again. Check that ${s}_{p}=10.30$ and that ${\tau }_{obs}=1.52$.
Two methods A and B, were used to find the latent heat of fusion of ice. 13 measurements were made with method A and 8 measurements were made with method B with statistics: ${\stackrel{̄}{x}}_{A}=80.02,\phantom{\rule{1em}{0ex}}{\stackrel{̄}{x}}_{B}=79.98,\phantom{\rule{1em}{0ex}}{s}_{A}=0.024$ and ${s}_{B}=0.031$. To test equal means assuming normality and equal variances, the observed test statistic (to 1dp) and the P-value are: Exactly one option must be correct)
 a) ${\tau }_{obs}=1.2;$ P-value = $P\left(|{t}_{19}|>1.2\right)>0.2$ b) ${\tau }_{obs}=3.3;$ P-value = $P\left({t}_{21}>3.3\right)<0.005$ c) ${\tau }_{obs}=3.3;$ P-value = $P\left({t}_{19}>3.3\right)<0.005$ d) ${\tau }_{obs}=3.3;$ P-value = $P\left(|{t}_{19}|>3.3\right)<0.01$

Choice (a) is incorrect
You have used a one-tail test and have also used the wrong $t$-distribution.
${s}_{p}=\sqrt{\frac{12×0.02{4}^{2}+7×0.03{1}^{2}}{13+8-2}}=0.02679$, so ${\tau }_{obs}=\frac{80.02-79.98}{0.02679\sqrt{\frac{1}{13}+\frac{1}{8}}}=3.3$,
with P-value = $P\left(|{t}_{19}|>3.3\right)=2P\left({t}_{19}>3.3\right)$, which has bounds (0.002, 0.01).