Use tables to find an interval containing
$P\left({t}_{6}>1.5\right)$. Exactly one
option must be correct)

*Choice (a) is incorrect*

Draw a diagram and mark in 1.5 on the x-axis. From the
tables, $P\left({t}_{6}>1.943\right)=0.05$.
Now mark in 1.943 on the x-axis, and it will be obvious that
$P\left({t}_{6}>1.5\right)$ must
be greater than 0.05.

*Choice (b) is correct!*

*Choice (c) is incorrect*

Draw a diagram and mark in 1.5 on the x-axis. From the tables,
$P\left({t}_{6}>1.440\right)=0.10$.
Now mark in 1.44 on the x-axis, and it will be obvious that
$P\left({t}_{6}>1.5\right)$ must
be smaller than 0.10.

*Choice (d) is incorrect*

Try again.

Let $\tau =\frac{\stackrel{\u0304}{X}-\mu}{S\u2215\sqrt{n}}$,
where $\stackrel{\u0304}{X}$
and $S$
are variables representing the mean and standard deviation of a sample of size
$n=13$ from the normal
$\mathcal{N}\left(\mu ,{\sigma}^{2}\right)$ population. From the
tables, the value of $c$
such that $P\left(\tau >c\right)=0.05$
is
Exactly one option must be correct)

*Choice (a) is incorrect*

Try again. $\tau $ does
not have the ${t}_{13}$
distribution.

*Choice (b) is incorrect*

Try
again. $\tau =\frac{\stackrel{\u0304}{X}-\mu}{S\u2215\sqrt{13}}$ is not normally
distributed.

*Choice (c) is incorrect*

What is
the distribution of $\tau =\frac{\stackrel{\u0304}{X}-\mu}{S\u2215\sqrt{13}}$?

*Choice (d) is correct!*

$\tau =\frac{\stackrel{\u0304}{X}-\mu}{S\u2215\sqrt{13}}\sim {t}_{12}$ and
$P\left({t}_{12}>1.782\right)=0.05$.

Questions 3-5 use the following information.

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units.

What are the null and alternative hypotheses being tested? Exactly one option must be correct)

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units.

What are the null and alternative hypotheses being tested? Exactly one option must be correct)

*Choice (a) is correct!*

This an upper-tail test, as the anticipated alternative hypothesis is a mean greater than 5 units.

*Choice (b) is incorrect*

Try again. What is the
anticipated alternative?

*Choice (c) is incorrect*

Try again. $\stackrel{\u0304}{x}$ is the
observed mean.

*Choice (d) is incorrect*

Try again. This is not a two-tailed test.

Questions 3-5 use the following information.

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units.

Assume that the measurement of fineness is normally distributed with a variance of ${\sigma}^{2}=4.0$. We can state: Exactly one option must be correct)

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units.

Assume that the measurement of fineness is normally distributed with a variance of ${\sigma}^{2}=4.0$. We can state: Exactly one option must be correct)

*Choice (a) is correct!*

Since
$\sigma =2$,
and this is an upper-tail test, the observed value of the Z-statistic is
$\frac{5.9-5}{2\u2215\sqrt{16}}=1.8$ with P-value
= $P\left(Z\ge 1.8\right)=0.0359$.

*Choice (b) is incorrect*

Try
again. The observed value of the Z-statistic is not 0.45.

*Choice (c) is incorrect*

Try
again. The observed value of the Z-statistic is not 3.6.

*Choice (d) is incorrect*

Try again. This is not
a two-tailed test.

Questions 3-5 use the following information.

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units. Assume that the measurement of fineness is normally distributed with unknown variance. State the distribution of the appropriate test statistic, $\tau $, and find the P-value if the standard deviation of the sample is $s=2.4$. Exactly one option must be correct)

The mean fineness $\mu $, of yarn is expected to be greater than the standard value of 5 units. To test this claim, the factory graded 16 specimens of the yarn and found the sample average, $\stackrel{\u0304}{x}$, to be 5.9 units. Assume that the measurement of fineness is normally distributed with unknown variance. State the distribution of the appropriate test statistic, $\tau $, and find the P-value if the standard deviation of the sample is $s=2.4$. Exactly one option must be correct)

*Choice (a) is incorrect*

Try again. The test statistic is not normal when
$\sigma $ is unknown.

*Choice (b) is incorrect*

Try again. You have
the wrong $t$ distribution.

*Choice (c) is incorrect*

Try again. You may
have misread the t-tables.

*Choice (d) is correct!*

${\tau}_{obs}=\frac{5.9-5}{2.4\u2215\sqrt{16}}=1.5$ and
$P\left({t}_{15}>1.5\right)$ is in
the interval (0.05, 0.10).

Questions 6-9 use the following information.

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

The management wishes to know if the new process results in a different mean
weight.
Which of the following are the most appropriate null and
alternative hypotheses? Exactly one option must be correct)

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

60.3 | 59.8 | 62.5 | 60.8 | 61.6 | 59.9 | 61.2 | 59.4 |

61.0 | 58.9 | 62.1 | 60.7 | 59.1 | 60.2 | 63.1 |

*Choice (a) is incorrect*

Try
again. There is no supporting information to anticipate an increase in weight.

*Choice (b) is correct!*

We
perform a two-sided test as there is no anticipated direction for the alternative hypothesis.

*Choice (c) is incorrect*

Try again. Think about the
value of $\mu $ and what sort of
test we need to perform.

*Choice (d) is incorrect*

Try again. Think
about the value of $\mu \phantom{\rule{0.3em}{0ex}}.$

Questions 6-9 use the following information.

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

The management wishes to know if the new process results in a different mean
weight.
Assuming the standard deviation is unchanged for the new process,
which of the following is the most appropriate test statistic,
$\tau $? Exactly one option
must be correct)

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

60.3 | 59.8 | 62.5 | 60.8 | 61.6 | 59.9 | 61.2 | 59.4 |

61.0 | 58.9 | 62.1 | 60.7 | 59.1 | 60.2 | 63.1 |

*Choice (a) is incorrect*

Try again, we have a sample of 15 measurements.

*Choice (b) is correct!*

Since
$\sigma $ is known, we
know that $\overline{X}\sim \mathcal{N}\left(\mu ,\frac{{\sigma}^{2}}{n}\right)\sim \phantom{\rule{0.3em}{0ex}}\mathcal{N}\left(60,\frac{1.{2}^{2}}{15}\right)$.
Standardising $\overline{X}$ gives
the test statistic, $\tau =\frac{\overline{X}-60}{1.2\u2215\sqrt{15}}\sim \mathcal{N}\left(0,1\right)$

*Choice (c) is incorrect*

You have not
standardised $\overline{X}$
correctly.

*Choice (d) is incorrect*

Try again, $\sigma $
is known.

Questions 6-9 use the following information.

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

The management wishes to know if the new process results in a different mean
weight.
State the P-value of the Z-test. Exactly one option must be correct)

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

60.3 | 59.8 | 62.5 | 60.8 | 61.6 | 59.9 | 61.2 | 59.4 |

61.0 | 58.9 | 62.1 | 60.7 | 59.1 | 60.2 | 63.1 |

*Choice (a) is incorrect*

Try again, remember we are performing a two-sided test.

*Choice (b) is incorrect*

Try again.

*Choice (c) is correct!*

Large and small values of
$\overline{X}$ argue
against ${H}_{0}^{}$.

The observed value of $\overline{X}$ is 60.71.

$P\left(\left|Z\right|>\frac{60.71-60}{1.2\u2215\sqrt{15}}\right)=P\left(\left|Z\right|>2.29\right)=2\left(1-\Phi \left(2.29\right)\right)=2\left(1-0.9890\right)=0.022.$

The observed value of $\overline{X}$ is 60.71.

$P\left(\left|Z\right|>\frac{60.71-60}{1.2\u2215\sqrt{15}}\right)=P\left(\left|Z\right|>2.29\right)=2\left(1-\Phi \left(2.29\right)\right)=2\left(1-0.9890\right)=0.022.$

*Choice (d) is incorrect*

Check that the sample mean is 60.71 and recalculate the observed
value of the Z-statistic.

Questions 6-9 use the following information.

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

The management wishes to know if the new process results in a different mean
weight.
If we are not prepared to assume that the standard deviation is 1.2g for the new process, find
${\tau}_{obs}$, the observed value of
the appropriate $t$-statistic,
and provide a P-value. Exactly one option must be correct)

A manufacturing process produces components which have weight normally distributed about 60g with a standard deviation of 1.2g. A new process has been developed to cut costs. Below are the weights of a random sample of 15 components produced by the new process:

60.3 | 59.8 | 62.5 | 60.8 | 61.6 | 59.9 | 61.2 | 59.4 |

61.0 | 58.9 | 62.1 | 60.7 | 59.1 | 60.2 | 63.1 |

*Choice (a) is incorrect*

Try again. You need to calculate both
$\stackrel{\u0304}{x}$ and
$s$ and substitute
into ${\tau}_{obs}=\frac{\stackrel{\u0304}{x}-60}{s\u2215\sqrt{15}}$.

*Choice (b) is incorrect*

Try again. Check your calculation of the sample standard deviation,
$s=1.24$, and
substitute $\stackrel{\u0304}{x}$
and $s$
into ${\tau}_{obs}=\frac{\stackrel{\u0304}{x}-60}{s\u2215\sqrt{15}}$.

*Choice (c) is incorrect*

Try again. This is not
an upper-tail test.

*Choice (d) is correct!*

Since $\sigma $ is
unknown, ${\tau}_{obs}=\frac{\overline{x}-\mu}{s\u2215\sqrt{n}}\phantom{\rule{0.3em}{0ex}}=\frac{60.71-60}{1.24\u2215\sqrt{15}}=2.22\phantom{\rule{0.3em}{0ex}}$

and P-value = $P\left(\left|{t}_{14}\right|>2.22\right)=2P\left({t}_{14}>2.22\right)<0.05$.

and P-value = $P\left(\left|{t}_{14}\right|>2.22\right)=2P\left({t}_{14}>2.22\right)<0.05$.

The breaking strengths of a certain brand of marine rope follow a normal distribution,
with unknown variance. To test whether the mean breaking strength is 8 units, 28
lengths of the rope were tested. From this sample, the observed value of the test statistic
was ${\tau}_{obs}=2.05$. The
appropriate $t$
distribution and approximate P-value are:
Exactly one option must be correct)

*Choice (a) is correct!*

The
P-value is $P\left(\left|{t}_{27}\right|>2.05\right)=2P\left({t}_{27}>2.05\right)\approx 2\times 0.025$.

*Choice (b) is incorrect*

Try again. Notice that we are conducting a two-sided test.

*Choice (c) is incorrect*

Try again. The
test statistic is not ${t}_{28}$.

*Choice (d) is incorrect*

Try again. The
test statistic is not ${t}_{8}$.