Question 1

$$\text { Using the } z \text { table, find the critical value (or values) for an } \alpha = 0.11 \text { right-tailed test. }$$

Accepted Answer

A)  1.60 
B)  0.80 
C)  0.62 
D)  1.23 
A)  1.60 
B)  0.80 
C)  0.62 
D)  1.23

Question 2

$$\text { Is the statement } H _ { 0 } : 18 = 6 \text { a valid null hypothesis? }$$

Accepted Answer

A)  No, there is no parameter contained in this statement. 
B)  Yes, this is a statement that compares two parameters. 
C)  No, equalities are not permitted in a null hypothesis. 
D)  Yes, this is a statement that compares a parameter to a value. 
A)  No, there is no parameter contained in this statement. 
B)  Yes, this is a statement that compares two parameters. 
C)  No, equalities are not permitted in a null hypothesis. 
D)  Yes, this is a statement that compares a parameter to a value.

Question 3

Assume that a $95 \%$ confidence interval for the mean is $10.5 < \mu < 15.0$. The null hypothesis $H _ { 0 } : \mu = 12.0$ at $\alpha = 0.05$ would

Accepted Answer

A)  not be rejected because 12.0 is less than 15.0. 
B)  be rejected because 12.0 is more than 10.5. 
C)  not be rejected because 12.0 is between 10.5 and 15.0. 
D)  be rejected because 12.0 is not equal to 12.75. 
A)  not be rejected because 12.0 is less than 15.0. 
B)  be rejected because 12.0 is more than 10.5. 
C)  not be rejected because 12.0 is between 10.5 and 15.0. 
D)  be rejected because 12.0 is not equal to 12.75.

Question 4

Which type of alternative hypothesis is used in the figure below?

Accepted Answer

A) $H _ { 1 } : \mu > k$ B) $H _ { 1 } : \mu = k$ C) $H _ { 1 } : \mu \neq k$ D) $H _ { 1 } : \mu < k$ A) $H _ { 1 } : \mu > k$ B) $H _ { 1 } : \mu = k$ C) $H _ { 1 } : \mu \neq k$ D) $H _ { 1 } : \mu < k$

Question 5

A test of $H _ { 0 } : \mu = 57$ versus $H _ { 1 } : \mu \neq 57$ is performed using a significance level of $\alpha = 0.01$. The value of the test statistic is $z = - 2.43$.
If the true value of $\mu$ is 55 , does the conclusion result in a Type I error, a Type II error, or correct decision?

Accepted Answer

A)  Type II error 
B)  Type I error 
C)  Correct decision 
A)  Type II error 
B)  Type I error 
C)  Correct decision

Question 6

Use technology to find the $P$-value for the following values of the test statistic $t$, sample size $n$, and alternate hypothesis $H _ { 1 }$.
$$t = 1.535 , n = 15 , H _ { 1 } : \mu \neq \mu _ { 0 }$$

Accepted Answer

A)  0.0735 
B)  0.1470 
C)  0.1456 
D)  0.0728 
A)  0.0735 
B)  0.1470 
C)  0.1456 
D)  0.0728

Question 7

A test of $H _ { 0 } : \mu = 59$ versus $H _ { 1 } : \mu < 59$ is performed using a significance level of $\alpha = 0.05$. The value of the test statistic is $z = - 1.80$.
If the true value of $\mu$ is 59 , does the conclusion result in a Type I error, a Type II error, or correct decision?

Accepted Answer

A)  Correct decision 
B)  Type II error 
C)  Type I error 
A)  Correct decision 
B)  Type II error 
C)  Type I error

Question 8

At a water bottling facility, a technician is testing a bottle filling machine that is supposed to deliver 750 milliliters of water. The technician dispenses 43 samples of water and determines the volume of each sample. The 43 samples have a mean volume of $\bar { x } = 751.1 \mathrm {~mL}$. The machine is out of calibration if the mean volume differs from $750 \mathrm {~mL}$.
The technician wants to perform a hypothesis test to determine whether the machine is ou calibration. The standard deviation of the dispensed volume is known to be $\sigma = 6.0$ value of the test statistic.

Accepted Answer

A)  1.20 
B)  0.17 
C)  2.94 
D)  0.18 
A)  1.20 
B)  0.17 
C)  2.94 
D)  0.18

Question 9

A statistician claims that the standard deviation of the weights of firemen is more than 25 pounds. A sample of 20 randomly chosen firemen had a standard deviation of their weights of $26.2$ pounds. Assume the variable is normally distributed. At $\alpha = 0.05$, what is the critical value $\chi ^ { 2 }$ for this test?

Accepted Answer

A)  10.851 
B)  10.117 
C)  31.410 
D)  30.144 
A)  10.851 
B)  10.117 
C)  31.410 
D)  30.144

Question 10

$$\text { Is the statement } H _ { 0 } : \mu = 6 \text { a valid null hypothesis? }$$

Accepted Answer

A)  No, equalities are not permitted in a null hypothesis. 
B)  Yes, this is a statement that compares a parameter to a value. 
C)  Yes, this is a statement that compares two parameters. 
D)  No, there is no parameter contained in this statement. 
A)  No, equalities are not permitted in a null hypothesis. 
B)  Yes, this is a statement that compares a parameter to a value. 
C)  Yes, this is a statement that compares two parameters. 
D)  No, there is no parameter contained in this statement.

Question 11

Historically, a certain region has experienced 65 thunder days annually. (A "thunder day" is day on which at least one instance of thunder is audible to a normal human ear). Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20. Can you conclude that the mean number of thunder days is less than 65 ? Use the $\alpha = 0.01$ level of significance.

Accepted Answer

A)  There is not enough information to draw a conclusion. 
B)  Yes. The number of thunder days appears to be less than 65. 
C)  No. There is insufficient evidence to conclude that the number of thunder days is less than 65. 
A)  There is not enough information to draw a conclusion. 
B)  Yes. The number of thunder days appears to be less than 65. 
C)  No. There is insufficient evidence to conclude that the number of thunder days is less than 65.

Question 12

About 31% of all burglaries are through an open or unlocked door or window. A sample of 137 burglaries indicated that 88 were not via an open or unlocked door or window. At
The 0.05 level of significance, can it be concluded that this differs from the stated
Proportion?

Accepted Answer

A)  There is not enough information to draw a conclusion. 
B)  Yes. There is enough evidence to support the claim that the proportion of open or unlocked window or door burglaries differs from 31%. 
C)  No. There is not enough evidence to support the claim that the proportion of open or unlocked window or door burglaries differs from 31%. 
A)  There is not enough information to draw a conclusion. 
B)  Yes. There is enough evidence to support the claim that the proportion of open or unlocked window or door burglaries differs from 31%. 
C)  No. There is not enough evidence to support the claim that the proportion of open or unlocked window or door burglaries differs from 31%.

Question 13

A lumber mill is tested for consistency by measuring the variance of board thickness. The target accuracy is a variance of $0.0035$ square inches or less. If 28 measurements are made and their variance is $0.006$ square inches, is there enough evidence to reject the claim that the standard deviation is within the limit at $\alpha = 0.01$ ?

Accepted Answer

A)  Yes, since the $\chi ^ { 2 }$ test value $8.91$ is less than the critical value 48.278. 
B)  Yes, since the $\chi ^ { 2 }$ test value $8.91$ is less than the critical value 46.963. 
C)  No, since the $\chi ^ { 2 }$ test value $46.29$ is less than the critical value $46.963$. 
D)  No, since the $\chi ^ { 2 }$ test value $46.29$ is less than the critical value $48.278$. 
A)  Yes, since the $\chi ^ { 2 }$ test value $8.91$ is less than the critical value 48.278. 
B)  Yes, since the $\chi ^ { 2 }$ test value $8.91$ is less than the critical value 46.963. 
C)  No, since the $\chi ^ { 2 }$ test value $46.29$ is less than the critical value $46.963$. 
D)  No, since the $\chi ^ { 2 }$ test value $46.29$ is less than the critical value $48.278$.

Question 14

The following output from MINITAB presents the results of a hypothesis test.
Test of $m u = 41$ vs. not $= 41$
The assumed standard deviation 11.9
$\begin{array}{llcccc}
\mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \text { CI } & \text { Z } & \text { P } \
42 & 38.23 & 2.16239 & (1.66073324,10.1373009) & -1.508542 & 0.131416 \
\hline
\end{array}$

Do you reject $H _ { 0 }$ at the $\alpha = 0.01$ level?

Accepted Answer

A)  Yes 
B)  No 
C)  There is not enough information to draw a conclusion. 
A)  Yes 
B)  No 
C)  There is not enough information to draw a conclusion.

Question 15

A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the sample must be less than or equal to $0.02$ ounce. A sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At $\alpha = 0.05$, can you reject the claim that the machine is functioning properly? Justify your answer. (A that the variables are approximately normally distributed.)
$$\begin{array} { l l l l } 
12.04 & 11.91 & 11.91 & 11.91 \
11.91 & 11.97 & 12.01 & 12.06
\end{array}$$

Accepted Answer

A)  $\chi ^ { 2 } = 72.000 , \chi _ { \text {critical } } ^ { 2 } = 14.067$; There is evidence to reject the claim that the machine is working properly. 
B)  $\chi ^ { 2 } = 65.570 , \chi _ { \text {critical } } ^ { 2 } = 15.507$; There is not enough evidence to reject the claim that the machine is working properly. 
C)  $\chi ^ { 2 } = 72.000 , \chi _ { \text {critical } } ^ { 2 } = 15.507$; There is evidence to reject the claim that the machine is working properly. 
D)  $\chi ^ { 2 } = 65.570 , \chi _ { \text {critical } } ^ { 2 } = 14.067$; There is evidence to reject the claim that the machine is working properly. 
A)  $\chi ^ { 2 } = 72.000 , \chi _ { \text {critical } } ^ { 2 } = 14.067$; There is evidence to reject the claim that the machine is working properly. 
B)  $\chi ^ { 2 } = 65.570 , \chi _ { \text {critical } } ^ { 2 } = 15.507$; There is not enough evidence to reject the claim that the machine is working properly. 
C)  $\chi ^ { 2 } = 72.000 , \chi _ { \text {critical } } ^ { 2 } = 15.507$; There is evidence to reject the claim that the machine is working properly. 
D)  $\chi ^ { 2 } = 65.570 , \chi _ { \text {critical } } ^ { 2 } = 14.067$; There is evidence to reject the claim that the machine is working properly.

Question 16

A test is made of $H _ { 0 } : \mu = 55$ versus $H _ { 1 } : \mu > 55 .$ A sample of size $n = 68$ is drawn, and $\bar { x } = 56$. The population standard deviation is $\sigma = 27$. Compute the value of the test statistic $z$.

Accepted Answer

A)  1.59 
B)  0.04 
C)  0.62 
D)  0.31 
A)  1.59 
B)  0.04 
C)  0.62 
D)  0.31

Question 17

The following output from MINITAB presents the results of a hypothesis test.
Test of $m u = 51$ vs. not $= 51$
The assumed standard deviation $12.2$

$\begin{array}{llcccc}
\mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} & \mathrm{Z} & \mathrm{P} \
49 & 54.30 & 2.240023 & (3.36669862,12.1475871) & 1.893443 & 0.058299
\end{array}$

What is the value of the test statistic?

Accepted Answer

A)  2.240023 
B)  0.058299 
C)  54.30 
D)  1.893443 
A)  2.240023 
B)  0.058299 
C)  54.30 
D)  1.893443

Question 18

The Golden Comet is a hybrid chicken that is prized for its high egg production rate and gentle disposition. According to recent studies, the mean rate of egg production for 1-year-old Golden Comets is $5.8$ eggs/week.

Sarah has 40 1-year-old hens that are fed exclusively on natural scratch feed: insects, seeds, and plants that the hens obtain as they range freely around the farm. Her hens exhibit a mean egg-laying rate of $6.2$ eggs/day.
Sarah wants to determine whether the mean laying rate $\mu$ for her hens is higher than the $\mathrm { m }$ rate for all Golden Comets. Assume the population standard deviation to be $\sigma = 1.7$ eggs/day. Compute the value of the test statistic.

Accepted Answer

A)  1.94 
B)  1.49 
C)  0.24 
D)  0.93 
A)  1.94 
B)  1.49 
C)  0.24 
D)  0.93

Question 19

A sample of 60 chewable vitamin tablets have a sample mean of 227 milligrams of vitamin C. Nutritionists want to perform a hypothesis test to determine how strong the
Evidence is that the mean mass of vitamin C per tablet exceeds 230 milligrams. State the
Appropriate null and alternate hypotheses.

Accepted Answer

A)  $H _ { 0 } : \mu  227$ 
B)  $H _ { 0 } : \mu = 230 , H _ { 1 } : \mu > 230$ 
C)  $H _ { 0 } : \mu > 227 , H _ { 1 } : \mu = 227$ 
D)  $H _ { 0 } : \mu = 230 , H _ { 1 } : \mu \neq 230$ 
A)  $H _ { 0 } : \mu  227$ 
B)  $H _ { 0 } : \mu = 230 , H _ { 1 } : \mu > 230$ 
C)  $H _ { 0 } : \mu > 227 , H _ { 1 } : \mu = 227$ 
D)  $H _ { 0 } : \mu = 230 , H _ { 1 } : \mu \neq 230$

Question 20

Dr. Christina Cuttleman, a nutritionist, claims that the average number of calories in a serving of popcorn is 75 with a standard deviation of 7 . A sample of 50 servings of popcorn was found to have an average of 78 calories. Check Dr. Cuttleman's claim at $\alpha = 0.05$.

Accepted Answer

@#LAT-DLM& (the claim) and @#LAT-DLM&
Critical val

$\text { Using the } z \text { table, find the critical value (or values) for an } \alpha = 0.11 \text { right-tailed test. }$

$\text { Is the statement } H _ { 0 } : 18 = 6 \text { a valid null hypothesis? }$

Assume that a $95 \%$ confidence interval for the mean is $10.5 < \mu < 15.0$ . The null hypothesis $H _ { 0 } : \mu = 12.0$ at $\alpha = 0.05$ would

Which type of alternative hypothesis is used in the figure below?

A test of $H _ { 0 } : \mu = 57$ versus $H _ { 1 } : \mu \neq 57$ is performed using a significance level of $\alpha = 0.01$ . The value of the test statistic is $z = - 2.43$ . If the true value of $\mu$ is 55 , does the conclusion result in a Type I error, a Type II error, or correct decision?

Use technology to find the $P$ -value for the following values of the test statistic $t$ , sample size $n$ , and alternate hypothesis $H _ { 1 }$ . $t = 1.535 , n = 15 , H _ { 1 } : \mu \neq \mu _ { 0 }$

A test of $H _ { 0 } : \mu = 59$ versus $H _ { 1 } : \mu < 59$ is performed using a significance level of $\alpha = 0.05$ . The value of the test statistic is $z = - 1.80$ . If the true value of $\mu$ is 59 , does the conclusion result in a Type I error, a Type II error, or correct decision?

$\text { Is the statement } H _ { 0 } : \mu = 6 \text { a valid null hypothesis? }$

About 31% of all burglaries are through an open or unlocked door or window. A sample of 137 burglaries indicated that 88 were not via an open or unlocked door or window. At The 0.05 level of significance, can it be concluded that this differs from the stated Proportion?

The following output from MINITAB presents the results of a hypothesis test. Test of $m u = 41$ vs. not $= 41$ The assumed standard deviation 11.9 Mean SE Mean 95\% CI Z P 42 38.23 2.16239 (1.66073324,10.1373009) -1.508542 0.131416 Do you reject $H _ { 0 }$ at the $\alpha = 0.01$ level?

A test is made of $H _ { 0 } : \mu = 55$ versus $H _ { 1 } : \mu > 55 .$ A sample of size $n = 68$ is drawn, and $\bar { x } = 56$ . The population standard deviation is $\sigma = 27$ . Compute the value of the test statistic $z$ .

The following output from MINITAB presents the results of a hypothesis test. Test of $m u = 51$ vs. not $= 51$ The assumed standard deviation $12.2$ Mean SE Mean 95\% 49 54.30 2.240023 (3.36669862,12.1475871) 1.893443 0.058299 What is the value of the test statistic?

A sample of 60 chewable vitamin tablets have a sample mean of 227 milligrams of vitamin C. Nutritionists want to perform a hypothesis test to determine how strong the Evidence is that the mean mass of vitamin C per tablet exceeds 230 milligrams. State the Appropriate null and alternate hypotheses.

Dr. Christina Cuttleman, a nutritionist, claims that the average number of calories in a serving of popcorn is 75 with a standard deviation of 7 . A sample of 50 servings of popcorn was found to have an average of 78 calories. Check Dr. Cuttleman's claim at $\alpha = 0.05$ .

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Probability and Counting Rules

Discrete Probability Distributions

The Normal Distribution

Confidence Intervals and Sample Size

Testing the Difference Between Two Means, Two Variances, and Two Proportions

Correlation and Regression

Other Chi-Square Tests

Analysis of Variance

Nonparametric Statistics

Sampling and Simulation

Filters

Exam 8: Hypothesis Testing

Is the statement H0:18=6 a valid null hypothesis? \text { Is the statement } H _ { 0 } : 18 = 6 \text { a valid null hypothesis? } Is the statement H0​:18=6 a valid null hypothesis?

Assume that a 95%95 \%95% confidence interval for the mean is 10.5<μ<15.010.5 < \mu < 15.010.5<μ<15.0 . The null hypothesis H0:μ=12.0H _ { 0 } : \mu = 12.0H0​:μ=12.0 at α=0.05\alpha = 0.05α=0.05 would

Which type of alternative hypothesis is used in the figure below?

Use technology to find the PPP -value for the following values of the test statistic ttt , sample size nnn , and alternate hypothesis H1H _ { 1 }H1​ . t=1.535,n=15,H1:μ≠μ0t = 1.535 , n = 15 , H _ { 1 } : \mu \neq \mu _ { 0 }t=1.535,n=15,H1​:μ=μ0​

Is the statement H0:μ=6 a valid null hypothesis? \text { Is the statement } H _ { 0 } : \mu = 6 \text { a valid null hypothesis? } Is the statement H0​:μ=6 a valid null hypothesis?

About 31% of all burglaries are through an open or unlocked door or window. A sample of 137 burglaries indicated that 88 were not via an open or unlocked door or window. At The 0.05 level of significance, can it be concluded that this differs from the stated Proportion?

A test is made of H0:μ=55H _ { 0 } : \mu = 55H0​:μ=55 versus H1:μ>55.H _ { 1 } : \mu > 55 .H1​:μ>55. A sample of size n=68n = 68n=68 is drawn, and xˉ=56\bar { x } = 56xˉ=56 . The population standard deviation is σ=27\sigma = 27σ=27 . Compute the value of the test statistic zzz .

The following output from MINITAB presents the results of a hypothesis test. Test of mu=51m u = 51mu=51 vs. not =51= 51=51 The assumed standard deviation 12.212.212.2 Mean SE Mean 95\% 49 54.30 2.240023 (3.36669862,12.1475871) 1.893443 0.058299 What is the value of the test statistic?

A sample of 60 chewable vitamin tablets have a sample mean of 227 milligrams of vitamin C. Nutritionists want to perform a hypothesis test to determine how strong the Evidence is that the mean mass of vitamin C per tablet exceeds 230 milligrams. State the Appropriate null and alternate hypotheses.

Dr. Christina Cuttleman, a nutritionist, claims that the average number of calories in a serving of popcorn is 75 with a standard deviation of 7 . A sample of 50 servings of popcorn was found to have an average of 78 calories. Check Dr. Cuttleman's claim at α=0.05\alpha = 0.05α=0.05 .

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Probability and Counting Rules

Discrete Probability Distributions

The Normal Distribution

Confidence Intervals and Sample Size

Testing the Difference Between Two Means, Two Variances, and Two Proportions

Correlation and Regression

Other Chi-Square Tests

Analysis of Variance

Nonparametric Statistics

Sampling and Simulation

Filters

$\text { Is the statement } H _ { 0 } : 18 = 6 \text { a valid null hypothesis? }$

Assume that a $95 \%$ confidence interval for the mean is $10.5 < \mu < 15.0$ . The null hypothesis $H _ { 0 } : \mu = 12.0$ at $\alpha = 0.05$ would

Use technology to find the $P$ -value for the following values of the test statistic $t$ , sample size $n$ , and alternate hypothesis $H _ { 1 }$ . $t = 1.535 , n = 15 , H _ { 1 } : \mu \neq \mu _ { 0 }$

$\text { Is the statement } H _ { 0 } : \mu = 6 \text { a valid null hypothesis? }$

A test is made of $H _ { 0 } : \mu = 55$ versus $H _ { 1 } : \mu > 55 .$ A sample of size $n = 68$ is drawn, and $\bar { x } = 56$ . The population standard deviation is $\sigma = 27$ . Compute the value of the test statistic $z$ .

The following output from MINITAB presents the results of a hypothesis test. Test of $m u = 51$ vs. not $= 51$ The assumed standard deviation $12.2$ Mean SE Mean 95\% 49 54.30 2.240023 (3.36669862,12.1475871) 1.893443 0.058299 What is the value of the test statistic?

Dr. Christina Cuttleman, a nutritionist, claims that the average number of calories in a serving of popcorn is 75 with a standard deviation of 7 . A sample of 50 servings of popcorn was found to have an average of 78 calories. Check Dr. Cuttleman's claim at $\alpha = 0.05$ .