Question 1

Determine whether the alternative hypothesis is left-tailed, right-tailed, or two-tailed.
$$H _ { 0 } : \mu = 71 \quad H _ { 1 } : \mu < 71$$

Accepted Answer

A)  left-tailed 
B)  right-tailed 
C)  two-tailed 
A)  left-tailed 
B)  right-tailed 
C)  two-tailed

Question 2

Doctors nationally believe that $70 \%$ of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At $\alpha = 0.05$ is this hospital's success rate different from the national average?

Accepted Answer

A)  No, because the test value 0.88 is in the noncritical region. 
B)  Yes, because the test value 0.88 is in the noncritical region. 
C)  No, because the test value 0.52 is in the noncritical region. 
D)  Yes, because the test value 0.52 is in the critical region. 
A)  No, because the test value 0.88 is in the noncritical region. 
B)  Yes, because the test value 0.88 is in the noncritical region. 
C)  No, because the test value 0.52 is in the noncritical region. 
D)  Yes, because the test value 0.52 is in the critical region.

Question 3

The mean annual tuition and fees for a sample of 12 private colleges was $27,900 with a standard deviation of $4400. A dotplot shows that it is reasonable to assume that the
Population is approximately normal. You wish to test whether the mean tuition and fees
For private colleges is different from $31,500.
State the null and alternate hypotheses.

Accepted Answer

A)  $H _ { 0 } : \mu \neq 31,500 , H _ { 1 } : \mu = 31,500$ 
B)  $H _ { 0 } : \mu = 31,500 , H _ { 1 } : \mu = 27,900$ 
C)  $H _ { 0 } : \mu = 27,900 , H _ { 1 } : \mu \neq 27,900$ 
D)  $H _ { 0 } : \mu = 31,500 , H _ { 1 } : \mu \neq 31,500$ 
A)  $H _ { 0 } : \mu \neq 31,500 , H _ { 1 } : \mu = 31,500$ 
B)  $H _ { 0 } : \mu = 31,500 , H _ { 1 } : \mu = 27,900$ 
C)  $H _ { 0 } : \mu = 27,900 , H _ { 1 } : \mu \neq 27,900$ 
D)  $H _ { 0 } : \mu = 31,500 , H _ { 1 } : \mu \neq 31,500$

Question 4

At a certain university, the average cost of books was $370 per student last semester and the population standard deviation was $80. This semester a sample of 45 students
Revealed an average cost of books of $390 per student. The Dean of Students believes
That the costs are greater this semester. What is the test value for this hypothesis?

Accepted Answer

A)  0.44 
B)  0.25 
C)  11.25 
D)  1.68 
A)  0.44 
B)  0.25 
C)  11.25 
D)  1.68

Question 5

A grocery store owner claims that the mean amount spent per checkout is more than $\$ 73$. A test is made of $H _ { 0 } : \mu = 73$ versus $H _ { 1 } : \mu > 73$. The null hypothesis is not rejected. State the appropriate conclusion.

Accepted Answer

A)  The mean checkout amount is greater than $73. 
B)  The mean checkout amount is less than or equal to $73. 
C)  There is not enough evidence to conclude that the mean checkout price is less than or equal to $73. 
D)  There is not enough evidence to conclude that the mean checkout price is greater than $73. 
A)  The mean checkout amount is greater than $73. 
B)  The mean checkout amount is less than or equal to $73. 
C)  There is not enough evidence to conclude that the mean checkout price is less than or equal to $73. 
D)  There is not enough evidence to conclude that the mean checkout price is greater than $73.

Question 6

The U.S. Bureau of Labor and Statistics reported that a person between the ages of 18 anc had an average of $9.3$ jobs. To see if this average is correct, a researcher selected a randor sample of 8 workers between the ages of 18 and 34 and asked how many different places worked. The results were as follows:
$\begin{array}{rrrr}
\hline 1 & 14 & 8 & 2 \
8 & 14 & 13 & 6 \
\hline
\end{array}$

At $\alpha = 0.05$, can it be concluded that the mean is $9.2 ?$ Use the $P$-value method. 34 has

Accepted Answer

A)  Yes. There is not enough evidence to reject the claim that the mean number of jobs is 9.2. 
B)  There is not enough information to draw a conclusion. 
C)  No. There is enough evidence to reject the claim that the mean number of jobs is 9.2. 
A)  Yes. There is not enough evidence to reject the claim that the mean number of jobs is 9.2. 
B)  There is not enough information to draw a conclusion. 
C)  No. There is enough evidence to reject the claim that the mean number of jobs is 9.2.

Question 7

A fleet of rental cars - all the same make, model, and year - has a mean fuel efficiency of $24.1$ miles per gallon (mpg). A random sample of 45 cars are selected and the air filter of each is replaced with a new one. Let $\mu$ be the population mean fuel efficiency score that would occur if every car's air filter were replaced. The air filter change is deemed effective if $\mu > 24.1 \mathrm { mpg }$. A test is made of $H _ { 0 } : \mu = 24.1$ versus $H _ { 1 } : \mu > 24.1$. Consider these possible conclusions:
i). The air filter changes are effective.
ii). The air filter changes are not effective.
iii). The air filter changes might not be successful.
Which of the three conclusions is best if $H _ { 0 }$ is rejected?

Accepted Answer

A)  iii 
B)  ii 
C)  $\mathrm { i }$ 
A)  iii 
B)  ii 
C)  $\mathrm { i }$

Question 8

If the null hypothesis $H _ { 0 } : \mu = 17.0$ is rejected at $\alpha = 0.10$ when a mean of $20.0$ is obtained from a random sample, one could also say that

Accepted Answer

A)  the 90% confidence interval for the population mean contains the value 20.0. 
B)  the 10% confidence interval for the population mean contains the value 20.0. 
C)  the 10% confidence interval for the population mean does not contain the value 17.0. 
D)  the 90% confidence interval for the population mean does not contain the value 17.0. 
A)  the 90% confidence interval for the population mean contains the value 20.0. 
B)  the 10% confidence interval for the population mean contains the value 20.0. 
C)  the 10% confidence interval for the population mean does not contain the value 17.0. 
D)  the 90% confidence interval for the population mean does not contain the value 17.0.

Question 9

A survey by Men's Health magazine stated that $14 \%$ of all men said that they used exercise to relieve stress. A random sample of 100 men was selected, and 9 said that they used exercise to relieve stress. Use the $P$-value method to test the claim. Use $\alpha = 0.10$.

Accepted Answer

A)  There is not enough information to draw a conclusion. 
B)  Yes. There is not enough evidence to reject the claim that the percentage of men who use exercise to relieve stress is 14%. 
C)  No. There is enough evidence to reject the claim that the percentage of men who use exercise to relieve stress is 14%. 
A)  There is not enough information to draw a conclusion. 
B)  Yes. There is not enough evidence to reject the claim that the percentage of men who use exercise to relieve stress is 14%. 
C)  No. There is enough evidence to reject the claim that the percentage of men who use exercise to relieve stress is 14%.

Question 10

The following display from a TI-84 Plus calculator presents the results of a hypothesis tes

What is the $P$-value?

Accepted Answer

A)  32 
B)  29.38 
C)  1.819327 
D)  -1.338686 
A)  32 
B)  29.38 
C)  1.819327 
D)  -1.338686

Question 11

The following output from MINITAB presents the results of a hypothesis test for a population mean $\mu$.
Test of mu $= 45$ vs. not $= 45$
$\begin{array}{lrrccrl}
\mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 95 \% \mathrm{CI} & \mathrm{T} & \mathrm{P} \
57 & 41.69 & 11.961865 & 1.584388 & (38.516105,44.863895) & -2.089135 & 0.0412
\end{array}$

How many degrees of freedom are there?

Accepted Answer

A)  58 
B)  45 
C)  56 
D)  57 
A)  58 
B)  45 
C)  56 
D)  57

Question 12

Use technology to find the $P$-value for the test value.
$$t = 1.532 , n = 22 \text {, right-tailed }$$

Accepted Answer

A)  0.1404 
B)  0.0696 
C)  0.0702 
D)  0.0351 
A)  0.1404 
B)  0.0696 
C)  0.0702 
D)  0.0351

Question 13

At a water bottling facility, a technician is testing a bottle filling machine that is supposed to deliver 500 milliliters of water. The technician dispenses 39 samples of water and determines the volume of each sample. The 39 samples have a mean volume of $\bar { x } = 498.3 \mathrm {~mL}$. The machine is out of calibration if the mean volume differs from $500 \mathrm {~mL}$.

The technician wants to perform a hypothesis test to determine whether the machi out of calibration. The standard deviation of the dispensed volume is known to be
i). State the appropriate null and alternate hypotheses.
ii). Compute the value of the test statistic.
iii). State a conclusion. Use the $\alpha = 0.05$ level of significance. s

Accepted Answer

i). @#LAT-DLM&
ii). @#LAT-DLM&
iii). Do not reject

Question 14

The following output from MINITAB presents the results of a hypothesis test.
Test of $m u = 40$ vs. not $= 40$
The assumed standard deviation $13.1$

$\begin{array}{llccrc}
\mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \text { CI } & \mathrm{Z} & \mathrm{P} \
61 & 43.33 & 1.575135 & (2.46057432,8.6351017) & 1.985354 & 0.047105 \
\hline
\end{array}$

What is the $P$-value?

Accepted Answer

A)  0.047105 
B)  43.33 
C)  1.985354 
D)  1.575135 
A)  0.047105 
B)  43.33 
C)  1.985354 
D)  1.575135

Question 15

The average monthly cell phone bill was reported to be $\$ 50.07$ by the U.S. Wireless Indus Random Sampling of a large cell phone company found the following monthly cell phone (in dollars):
$\begin{array}{llll}
\hline 52.62 & 46.27 & 59.77 & 67.21 \
57.26 & 49.24 & 45.99 & 46.81 \
55.39 & 48.08 & & \
\hline
\end{array}$

At the $0.05$ level of significance, can it be concluded that the average phone bill has incre ry. reased?

Accepted Answer

A)  No. There is not enough evidence to support the claim that the average bill has increased from $50.07. 
B)  There is not enough information to draw a conclusion. 
C)  Yes. There is enough evidence to support the claim that the average bill has increased from $50.07. 
A)  No. There is not enough evidence to support the claim that the average bill has increased from $50.07. 
B)  There is not enough information to draw a conclusion. 
C)  Yes. There is enough evidence to support the claim that the average bill has increased from $50.07.

Question 16

$$\text { A type II error occurs if one does not reject the null hypothesis when it is }$$__________

Accepted Answer

A) True 
 B)False

Question 17

A random sample of stipends of teaching assistants in economics is listed. Is there sufficient evidence at the $\alpha = 0.05$ level to conclude that the average stipend differs from $\$ 15,000$ ? The stipends listed (in dollars) are for the academic year.
$$\begin{array} { l l l l } 
\hline 13,500 & 18,500 & 11,500 & 14,856 \
12,685 & 13,919 & 13,500 & 12,481 \
17,104 & 12,783 & 15,838 & 15,500 \
\hline
\end{array}$$

Accepted Answer

A)  There is not enough information to draw a conclusion. 
B)  No. There is not enough evidence to conclude that the average stipend differs from $15,000. 
C)  Yes. There is enough evidence to conclude that the average stipend differs from $15,000. 
A)  There is not enough information to draw a conclusion. 
B)  No. There is not enough evidence to conclude that the average stipend differs from $15,000. 
C)  Yes. There is enough evidence to conclude that the average stipend differs from $15,000.

Question 18

A test is made of $H _ { 0 } : \mu = 54$ versus $H _ { 1 } : \mu \neq 54$. A sample of size $n = 73$ is drawn, and $\bar { x } = 57$. The population standard deviation is $\sigma = 21$. Compute the value of the test statistic $z$.

Accepted Answer

A)  5.59 
B)  0.89 
C)  1.22 
D)  0.14 
A)  5.59 
B)  0.89 
C)  1.22 
D)  0.14

Question 19

State the appropriate null and alternative hypothesis and find the critical value for a right-tailed test with $\alpha = 0.05$ and $n = 18$. Use $\sigma ^ { 2 } = 256$.

Accepted Answer

A) $\mathrm { H } _ { \mathrm { O } } : \sigma 2 = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } > 256$ C. $\mathrm { V } . = 27.587$ B) $\mathrm { H } _ { \mathrm { O } : } \sigma ^ { 2 } = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } > 256$ C. $\mathrm { V } . = 28.869$ C) $\mathrm { H } _ { \mathrm { O } } : \sigma 2 = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } < 256$ C. $\mathrm { V } . = 27.587$ D) $\mathrm { H } _ { \mathrm { O } } : \sigma ^ { 2 } \neq 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } < 256$ C. $\mathrm { V } . = 28.869$ A) $\mathrm { H } _ { \mathrm { O } } : \sigma 2 = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } > 256$ C. $\mathrm { V } . = 27.587$ B) $\mathrm { H } _ { \mathrm { O } : } \sigma ^ { 2 } = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } > 256$ C. $\mathrm { V } . = 28.869$ C) $\mathrm { H } _ { \mathrm { O } } : \sigma 2 = 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } < 256$ C. $\mathrm { V } . = 27.587$ D) $\mathrm { H } _ { \mathrm { O } } : \sigma ^ { 2 } \neq 256$ $\mathrm { H } _ { 1 : } \sigma ^ { 2 } < 256$ C. $\mathrm { V } . = 28.869$

Question 20

A report by the Gallup Poll stated that on average a woman contacts her physician $5.8$ tim year. A researcher randomly selects 20 women and obtained these data.

At $\alpha = 0.05$, can it be concluded that the average is still $5.8$ visits per year? Use the $P$-value method. s a

Accepted Answer

A)  No. There is enough evidence to reject the claim that the mean number of vists per year is 5.8. 
B)  There is not enough information to draw a conclusion. 
C)  Yes. There is not enough evidence to reject the claim that the mean number of vists per year is 5.8. 
A)  No. There is enough evidence to reject the claim that the mean number of vists per year is 5.8. 
B)  There is not enough information to draw a conclusion. 
C)  Yes. There is not enough evidence to reject the claim that the mean number of vists per year is 5.8.

Determine whether the alternative hypothesis is left-tailed, right-tailed, or two-tailed. $H _ { 0 } : \mu = 71 \quad H _ { 1 } : \mu < 71$

Doctors nationally believe that $70 \%$ of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At $\alpha = 0.05$ is this hospital's success rate different from the national average?

A grocery store owner claims that the mean amount spent per checkout is more than $\$ 73$ . A test is made of $H _ { 0 } : \mu = 73$ versus $H _ { 1 } : \mu > 73$ . The null hypothesis is not rejected. State the appropriate conclusion.

If the null hypothesis $H _ { 0 } : \mu = 17.0$ is rejected at $\alpha = 0.10$ when a mean of $20.0$ is obtained from a random sample, one could also say that

A survey by Men's Health magazine stated that $14 \%$ of all men said that they used exercise to relieve stress. A random sample of 100 men was selected, and 9 said that they used exercise to relieve stress. Use the $P$ -value method to test the claim. Use $\alpha = 0.10$ .

The following display from a TI-84 Plus calculator presents the results of a hypothesis tes What is the $P$ -value?

The following output from MINITAB presents the results of a hypothesis test for a population mean $\mu$ . Test of mu $= 45$ vs. not $= 45$ Mean StDev SE Mean 95\% 57 41.69 11.961865 1.584388 (38.516105,44.863895) -2.089135 0.0412 How many degrees of freedom are there?

Use technology to find the $P$ -value for the test value. $t = 1.532 , n = 22 \text {, right-tailed }$

The following output from MINITAB presents the results of a hypothesis test. Test of $m u = 40$ vs. not $= 40$ The assumed standard deviation $13.1$ Mean SE Mean 95\% CI 61 43.33 1.575135 (2.46057432,8.6351017) 1.985354 0.047105 What is the $P$ -value?

$\text { A type II error occurs if one does not reject the null hypothesis when it is }$ __________

A test is made of $H _ { 0 } : \mu = 54$ versus $H _ { 1 } : \mu \neq 54$ . A sample of size $n = 73$ is drawn, and $\bar { x } = 57$ . The population standard deviation is $\sigma = 21$ . Compute the value of the test statistic $z$ .

State the appropriate null and alternative hypothesis and find the critical value for a right-tailed test with $\alpha = 0.05$ and $n = 18$ . Use $\sigma ^ { 2 } = 256$ .

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

Determine whether the alternative hypothesis is left-tailed, right-tailed, or two-tailed. H0:μ=71H1:μ<71H _ { 0 } : \mu = 71 \quad H _ { 1 } : \mu < 71H0​:μ=71H1​:μ<71

Doctors nationally believe that 70%70 \%70% of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At α=0.05\alpha = 0.05α=0.05 is this hospital's success rate different from the national average?

A grocery store owner claims that the mean amount spent per checkout is more than $73\$ 73$73 . A test is made of H0:μ=73H _ { 0 } : \mu = 73H0​:μ=73 versus H1:μ>73H _ { 1 } : \mu > 73H1​:μ>73 . The null hypothesis is not rejected. State the appropriate conclusion.

If the null hypothesis H0:μ=17.0H _ { 0 } : \mu = 17.0H0​:μ=17.0 is rejected at α=0.10\alpha = 0.10α=0.10 when a mean of 20.020.020.0 is obtained from a random sample, one could also say that

A survey by Men's Health magazine stated that 14%14 \%14% of all men said that they used exercise to relieve stress. A random sample of 100 men was selected, and 9 said that they used exercise to relieve stress. Use the PPP -value method to test the claim. Use α=0.10\alpha = 0.10α=0.10 .

The following display from a TI-84 Plus calculator presents the results of a hypothesis tes What is the PPP -value?

The following output from MINITAB presents the results of a hypothesis test for a population mean μ\muμ . Test of mu =45= 45=45 vs. not =45= 45=45 Mean StDev SE Mean 95\% 57 41.69 11.961865 1.584388 (38.516105,44.863895) -2.089135 0.0412 How many degrees of freedom are there?

Use technology to find the PPP -value for the test value. t=1.532,n=22, right-tailed t = 1.532 , n = 22 \text {, right-tailed }t=1.532,n=22, right-tailed

The following output from MINITAB presents the results of a hypothesis test. Test of mu=40m u = 40mu=40 vs. not =40= 40=40 The assumed standard deviation 13.113.113.1 Mean SE Mean 95\% CI 61 43.33 1.575135 (2.46057432,8.6351017) 1.985354 0.047105 What is the PPP -value?

A type II error occurs if one does not reject the null hypothesis when it is \text { A type II error occurs if one does not reject the null hypothesis when it is } A type II error occurs if one does not reject the null hypothesis when it is __________

A test is made of H0:μ=54H _ { 0 } : \mu = 54H0​:μ=54 versus H1:μ≠54H _ { 1 } : \mu \neq 54H1​:μ=54 . A sample of size n=73n = 73n=73 is drawn, and xˉ=57\bar { x } = 57xˉ=57 . The population standard deviation is σ=21\sigma = 21σ=21 . Compute the value of the test statistic zzz .

State the appropriate null and alternative hypothesis and find the critical value for a right-tailed test with α=0.05\alpha = 0.05α=0.05 and n=18n = 18n=18 . Use σ2=256\sigma ^ { 2 } = 256σ2=256 .

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

Determine whether the alternative hypothesis is left-tailed, right-tailed, or two-tailed. $H _ { 0 } : \mu = 71 \quad H _ { 1 } : \mu < 71$

Doctors nationally believe that $70 \%$ of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At $\alpha = 0.05$ is this hospital's success rate different from the national average?

A grocery store owner claims that the mean amount spent per checkout is more than $\$ 73$ . A test is made of $H _ { 0 } : \mu = 73$ versus $H _ { 1 } : \mu > 73$ . The null hypothesis is not rejected. State the appropriate conclusion.

If the null hypothesis $H _ { 0 } : \mu = 17.0$ is rejected at $\alpha = 0.10$ when a mean of $20.0$ is obtained from a random sample, one could also say that

A survey by Men's Health magazine stated that $14 \%$ of all men said that they used exercise to relieve stress. A random sample of 100 men was selected, and 9 said that they used exercise to relieve stress. Use the $P$ -value method to test the claim. Use $\alpha = 0.10$ .

The following display from a TI-84 Plus calculator presents the results of a hypothesis tes What is the $P$ -value?

The following output from MINITAB presents the results of a hypothesis test for a population mean $\mu$ . Test of mu $= 45$ vs. not $= 45$ Mean StDev SE Mean 95\% 57 41.69 11.961865 1.584388 (38.516105,44.863895) -2.089135 0.0412 How many degrees of freedom are there?

Use technology to find the $P$ -value for the test value. $t = 1.532 , n = 22 \text {, right-tailed }$

The following output from MINITAB presents the results of a hypothesis test. Test of $m u = 40$ vs. not $= 40$ The assumed standard deviation $13.1$ Mean SE Mean 95\% CI 61 43.33 1.575135 (2.46057432,8.6351017) 1.985354 0.047105 What is the $P$ -value?

$\text { A type II error occurs if one does not reject the null hypothesis when it is }$ __________

A test is made of $H _ { 0 } : \mu = 54$ versus $H _ { 1 } : \mu \neq 54$ . A sample of size $n = 73$ is drawn, and $\bar { x } = 57$ . The population standard deviation is $\sigma = 21$ . Compute the value of the test statistic $z$ .

State the appropriate null and alternative hypothesis and find the critical value for a right-tailed test with $\alpha = 0.05$ and $n = 18$ . Use $\sigma ^ { 2 } = 256$ .