## MATH1015 Quizzes

Quiz 9: t-tables; Z-tests and t-tests for a mean
Question 1 Questions
Use tables to find an interval containing $P\left({t}_{6}>1.5\right)$. Exactly one option must be correct)
 a) (0, 0.05) b) (0.05, 0.10) c) (0.10, 1) d) (0.25, 1)

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 $\tau =\frac{\stackrel{̄}{X}-\mu }{S∕\sqrt{n}}$, where $\stackrel{̄}{X}$ and $S$ are variables representing the mean and standard deviation of a sample of size $n=13$ from the normal $\mathsc{N}\left(\mu ,{\sigma }^{2}\right)$ population. From the tables, the value of $c$ such that $P\left(\tau >c\right)=0.05$ is Exactly one option must be correct)
 a) $c=1.771$ b) $c=1.645$ c) $c=0.05$ d) $c=1.782$

Choice (a) is incorrect
Try again. $\tau$ does not have the ${t}_{13}$ distribution.
Choice (b) is incorrect
Try again. $\tau =\frac{\stackrel{̄}{X}-\mu }{S∕\sqrt{13}}$ is not normally distributed.
Choice (c) is incorrect
What is the distribution of $\tau =\frac{\stackrel{̄}{X}-\mu }{S∕\sqrt{13}}$?
Choice (d) is correct!
$\tau =\frac{\stackrel{̄}{X}-\mu }{S∕\sqrt{13}}\sim {t}_{12}$ 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, $\stackrel{̄}{x}$, to be 5.9 units.
What are the null and alternative hypotheses being tested? Exactly one option must be correct)
 a) ${H}_{0}:\mu =5;\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{H}_{1}:\mu >5$ b) ${H}_{0}:\mu =5;\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{H}_{1}:\mu =5.9$ c) ${H}_{0}:\stackrel{̄}{x}=5;\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{H}_{1}:\stackrel{̄}{x}>5.9$ d) ${H}_{0}:\mu =5;\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{H}_{1}:\mu \ne 5$

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. $\stackrel{̄}{x}$ 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, $\stackrel{̄}{x}$, 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)
 a) The P-value is $P\left(Z\ge 1.8\right)=0.0359$ and there is strong evidence that the mean exceeds 5 units. b) The P-value is $P\left(Z\ge 0.45\right)=0.3264$ which indicates the data are consistent with a mean of 5 units. c) The P-value is $P\left(Z\ge 3.6\right)=0.0002$ and there is strong evidence that the mean exceeds 5 units. d) The P-value is $P\left(|Z|\ge 1.8\right)=0.0718$ which does not constitute strong evidence against a mean of 5 units.

Choice (a) is correct!
Since $\sigma =2$, and this is an upper-tail test, the observed value of the Z-statistic is $\frac{5.9-5}{2∕\sqrt{16}}=1.8$ 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, $\stackrel{̄}{x}$, 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)
 a) $\tau \sim \mathsc{N}\left(0,1\right)$ and P-value $<0.05$. b) $\tau \sim {t}_{16}$ and P-value $<0.05$. c) $\tau \sim {t}_{15}$ and P-value $<0.05$. d) $\tau \sim {t}_{15}$ and P-value $>0.05$.

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!
${\tau }_{obs}=\frac{5.9-5}{2.4∕\sqrt{16}}=1.5$ 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:
 60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4 61 58.9 62.1 60.7 59.1 60.2 63.1
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) ${H}_{0}^{}:\mu =60$; ${H}_{1}^{}:\mu >60$ b) ${H}_{0}^{}:\mu =60$; ${H}_{1}^{}:\mu \ne 60$ c) ${H}_{0}^{}:\mu =0$; ${H}_{1}^{}:\mu >0$ d) ${H}_{0}^{}:\mu =0$; ${H}_{1}^{}:\mu \ne 0$

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:
 60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4 61 58.9 62.1 60.7 59.1 60.2 63.1
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) $\tau =X\sim \mathsc{N}\left(60,1.{2}^{2}\right)$ b) $\tau =\frac{\overline{X}-60}{1.2∕\sqrt{15}}\sim \mathsc{N}\left(0,1\right)$ c) $\tau =\frac{\overline{X}-60}{1.2}\sim \mathsc{N}\left(0,1\right)$ d) $\tau =\frac{\overline{X}-\mu }{S∕\sqrt{n}}\sim {t}_{14}$

Choice (a) is incorrect
Try again, we have a sample of 15 measurements.
Choice (b) is correct!
Since $\sigma$ is known, we know that $\overline{X}\sim \mathsc{N}\left(\mu ,\frac{{\sigma }^{2}}{n}\right)\sim \phantom{\rule{0.3em}{0ex}}\mathsc{N}\left(60,\frac{1.{2}^{2}}{15}\right)$. Standardising $\overline{X}$ gives the test statistic, $\tau =\frac{\overline{X}-60}{1.2∕\sqrt{15}}\sim \mathsc{N}\left(0,1\right)$
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:
 60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4 61 58.9 62.1 60.7 59.1 60.2 63.1
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) P-value = 0.011. b) P-value = 0.989. c) P-value = 0.022. d) P-value = 0.10

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.
$P\left(|Z|>\frac{60.71-60}{1.2∕\sqrt{15}}\right)=P\left(|Z|>2.29\right)=2\left(1-\Phi \left(2.29\right)\right)=2\left(1-0.9890\right)=0.022.$
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:
 60.3 59.8 62.5 60.8 61.6 59.9 61.2 59.4 61 58.9 62.1 60.7 59.1 60.2 63.1
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) ${\tau }_{obs}=60.71$, P-value = 0. b) ${\tau }_{obs}=1.79$, P-value =$P\left(|{t}_{14}|>1.79\right)>0.05$. c) ${\tau }_{obs}=2.22$, P-value = $P\left({t}_{14}>2.22\right)<0.025$. d) ${\tau }_{obs}=2.22$, P-value = $P\left(|{t}_{14}|>2.22\right)<0.05$.

Choice (a) is incorrect
Try again. You need to calculate both $\stackrel{̄}{x}$ and $s$ and substitute into ${\tau }_{obs}=\frac{\stackrel{̄}{x}-60}{s∕\sqrt{15}}$.
Choice (b) is incorrect
Try again. Check your calculation of the sample standard deviation, $s=1.24$, and substitute $\stackrel{̄}{x}$ and $s$ into ${\tau }_{obs}=\frac{\stackrel{̄}{x}-60}{s∕\sqrt{15}}$.
Choice (c) is incorrect
Try again. This is not an upper-tail test.
Choice (d) is correct!
Since $\sigma$ is unknown, ${\tau }_{obs}=\frac{\overline{x}-\mu }{s∕\sqrt{n}}\phantom{\rule{0.3em}{0ex}}=\frac{60.71-60}{1.24∕\sqrt{15}}=2.22\phantom{\rule{0.3em}{0ex}}$
and P-value = $P\left(|{t}_{14}|>2.22\right)=2P\left({t}_{14}>2.22\right)<0.05$.
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)
 a) ${t}_{27}$ , P-value $\approx 0.05$. b) ${t}_{27}$ , P-value $\approx 0.025$. c) ${t}_{28}$ , P-value $\approx 0.05$. d) ${t}_{8}$ , P-value $\approx 0.025$.

Choice (a) is correct!
The P-value is $P\left(|{t}_{27}|>2.05\right)=2P\left({t}_{27}>2.05\right)\approx 2×0.025$.
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 ${t}_{8}$.