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)

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)

*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)

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)

*Choice (a) is correct!*

Write
$\tau $
to represent the number of weight increases. Under
${H}_{0}$,
$\tau \sim \mathcal{\mathcal{B}}\left(11,0.5\right)$. 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 $\stackrel{\u0304}{x}=-0.6$ 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)

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 $\stackrel{\u0304}{x}=-0.6$ 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)

*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!*

${\tau}_{obs}=\frac{-0.6}{1.2\u2215\sqrt{11}}=-1.66$, 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)

${d}_{i}$: 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)

*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 ${\tau}_{obs}=\frac{1.325}{2.536\u2215\sqrt{12}}=1.809$.
The test is two-sided, so the P-value =
$P\left(\left|{t}_{11}\right|>1.809\right)=2P\left({t}_{11}>1.809\right)$,
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)

*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,
$X\sim \mathcal{\mathcal{B}}\left(64,0.5\right)$ 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 $Y\sim \mathcal{N}\left(32,16\right)$
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)

*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 =
$P\left(\left|{t}_{20}\right|\ge 1.95\right)=2P\left({t}_{20}\ge 1.95\right)$. 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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

The pooled estimate of the standard deviation, ${s}_{p}$, in the two-sample test statistic is: Exactly one option must be correct)

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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

The pooled estimate of the standard deviation, ${s}_{p}$, in the two-sample test statistic is: Exactly one option must be correct)

*Choice (a) is correct!*

${s}_{p}=\sqrt{\frac{9\times 9.1{5}^{2}+13\times 11.0{2}^{2}}{10+14-2}}=10.30$.

*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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

To test for equal means, the appropriate $t$-distribution is Exactly one option must be correct)

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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

To test for equal means, the appropriate $t$-distribution is Exactly one option must be correct)

*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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

Assuming the two populations sampled are normal with equal variance, find the P-value. Exactly one option must be correct)

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: $\stackrel{\u0304}{x}=129.44$, $\u0233=122.65$, ${s}_{x}=9.15$, ${s}_{y}=11.02$.

Assuming the two populations sampled are normal with equal variance, find the P-value. Exactly one option must be correct)

*Choice (a) is incorrect*

This is a two-sided test.

*Choice (b) is correct!*

${s}_{p}=\sqrt{\frac{9\times 9.1{5}^{2}+13\times 11.0{2}^{2}}{10+14-2}}=10.30$, so
${\tau}_{obs}=\frac{129.44-122.65}{10.30\sqrt{\frac{1}{10}+\frac{1}{14}}}=1.52$.

P-value = $P\left(\left|{t}_{22}\right|\ge 1.52\right)=2P\left({t}_{22}\ge 1.52\right)$, with bounds (0.1, 0.2), since $P\left({t}_{22}\ge 1.52\right)$ is in the range (0.05, 0.10).

P-value = $P\left(\left|{t}_{22}\right|\ge 1.52\right)=2P\left({t}_{22}\ge 1.52\right)$, 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:
${\stackrel{\u0304}{x}}_{A}=80.02,\phantom{\rule{1em}{0ex}}{\stackrel{\u0304}{x}}_{B}=79.98,\phantom{\rule{1em}{0ex}}{s}_{A}=0.024$ 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)

*Choice (a) is incorrect*

Recalculate your
test statistic.

*Choice (b) is incorrect*

You have used a one-tail test and have also used the wrong
$t$-distribution.

*Choice (c) is incorrect*

This should be
a two-tail test

*Choice (d) is correct!*

${s}_{p}=\sqrt{\frac{12\times 0.02{4}^{2}+7\times 0.03{1}^{2}}{13+8-2}}=0.02679$, so
${\tau}_{obs}=\frac{80.02-79.98}{0.02679\sqrt{\frac{1}{13}+\frac{1}{8}}}=3.3$,

with P-value = $P\left(\left|{t}_{19}\right|>3.3\right)=2P\left({t}_{19}>3.3\right)$, which has bounds (0.002, 0.01).

with P-value = $P\left(\left|{t}_{19}\right|>3.3\right)=2P\left({t}_{19}>3.3\right)$, which has bounds (0.002, 0.01).