Question 1

Based on data from two very large independent samples, two students tested a hypothesis about
Equality of population means using $$\alpha = 0.02$$
2) One student used a one-tail test and rejected the null
Hypothesis, but the other used a two-tail test and failed to reject the null. Which of these might
Have been their calculated value of t?
A) 2.22
B) 1.88
C) 1.22
D) 2.66
E) 1.55

Accepted Answer

For a one-tail test at $\alpha = 0.02$, the critical value is lower than for a two-tail test. A t-value of 2.22 could be significant in a one-tail test but not in a two-tail test, leading to rejection in the former and not in the latter.

Question 2

You could win a $1000 prize by tossing a coin in one of two games. To win Game A, you must get
Exactly 50% heads. To win Game B, you must get between 45% and 55% heads. Although which
Game you must play will be chosen randomly, then you may decide whether to toss the coin 20
Times or 50 times. How many tosses would you choose to make?
A) It does not matter.
B) 50 tosses for A, 20 tosses for B
C) 50 tosses for either game
D) 20 tosses for A, 50 tosses for B
E) 20 tosses for either game

Accepted Answer

For Game A, with 20 tosses, getting exactly 50% heads (10 heads) is more likely than with 50 tosses (exactly 25 heads). For Game B, 50 tosses increases the probability of getting between 45% and 55% heads compared to 20 tosses.

Question 3

An elementary school principal wants to know the mean number of children in families whose
Children attend this school. He checks all the families using the school's registration records, and we
Create a 95% confidence interval based on a t-distribution. This procedure was not appropriate.
Why?
A) At a given school families are not randomly selected.
B) The population standard deviation is known, so he should have used a z-model.
C) The entire population of families was gathered so there is no reason to do inference.
D) Since these families are from only one school, the family sizes may be skewed.
E) The recent record-setting family with twelve children is probably an outlier.

Accepted Answer

C

Question 4

At one SAT test site students taking the test for a second time volunteered to inhale supplemental
Oxygen for 10 minutes before the test. In fact, some received oxygen, but others (randomly
Assigned) were given just normal air. Test results showed that 42 of 66 students who breathed
Oxygen improved their SAT scores, compared to only 35 of 63 students who did not get the oxygen.
Which procedure should we use to see if there is evidence that breathing extra oxygen can help
Test-takers think more clearly? 
A) 2-proportion $z$-test
B) 1 -proportion z-test
C) 1 -sample $t$-test
D) matched pairs $t$-test
E) 2-sample $t$-test

Accepted Answer

The correct procedure is a 2-proportion z-test because we are comparing the proportions of students who improved their SAT scores in two independent groups: those who breathed oxygen and those who did not.

Question 5

A marketing company reviewing the length of television commercials monitored a random sample
Of commercials over several days. They found that a 95% confidence interval for the mean length
(in seconds) of commercials aired daily was (23, 27). Which is true?
A) 95% of the commercials they checked were between 23 and 27 seconds long.
B) Commercials average between 23 and 27 seconds long on 95% of the days.
C) 95% of all the commercials aired were between 23 and 27 seconds a day.
D) 95% of all samples would show mean commercial length between 23 and 27 seconds.
E) We're 95% sure that the mean commercial length is between 23 and 27 seconds.

Accepted Answer

A 95% confidence interval means we are 95% confident that the true mean length of commercials is between 23 and 27 seconds.

Question 6

A professor was curious about her students' grade point averages (GPAs). She took a random
Sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of
The following formulas gives a 99% confidence interval for the mean GPA of the professor's
Students? 
A) $3.01 \pm 2.977 ( 0.534 / \sqrt { 15 } )$
B) $3.01 \pm 2.576 ( 0.534 / \sqrt { 15 } )$
C) $3.01 \pm 2.977 ( 0.534 / \sqrt { 14 } )$
D) $3.01 \pm 2.947 ( 0.534 / \sqrt { 15 } )$
E) $3.01 \pm 2.947 ( 0.534 / \sqrt { 14 } )$

Accepted Answer

The correct formula for a 99% confidence interval for the mean GPA is:$ \mu \pm z_{\alpha/2} ( \sigma / \sqrt { n } ) $where:$\mu$ is the population mean$z_{\alpha/2}$ is the z-score corresponding to the desired confidence level$\sigma$ is the population standard deviation$n$ is the sample sizeIn this case, we have:$\mu = 3.01$$z_{\alpha/2} = 2.977$ (for a 99% confidence level)$\sigma = 0.534$$n = 15$Plugging these values into the formula, we get:$3.01 \pm 2.977 ( 0.534 / \sqrt { 15 } )$Therefore, the correct answer is A.

Question 7

A survey asked people "On what percent of days do you get more than 30 minutes of vigorous
Exercise?" Using their responses we want to estimate the difference in exercise frequency between
Men and women. We should use a
A) matched pairs t-interval
B) 1-proportion z-interval
C) 2-sample t-interval
D) 1-sample t-interval
E) 2-proportion z-interval

Accepted Answer

A 2-sample t-interval is appropriate for estimating the difference in means between two independent groups, such as men and women, based on their responses.

Question 8

We want to know the mean winning score at the US Open golf championship. An internet search
Gives us all the scores for the history of that tournament, and we create a 95% confidence interval
Based on a t-distribution. This procedure was not appropriate. Why?
A) The population standard deviation is known, so we should have used a z-model.
B) Since these are the best players in the world, the scores are probably skewed.
C) The entire population of scores was gathered so there is no reason to do inference.
D) In big golf tournaments the players are not randomly selected.
E) The recent record-setting score is probably an outlier.

Accepted Answer

Since the entire population of scores was gathered, there is no need to perform inference, as inference is used to estimate population parameters from a sample.

Question 9

Absorption rates into the body are important considerations when manufacturing a generic version
Of a brand-name drug. A pharmacist read that the absorption rate into the body of a new generic
Drug (G) is the same as its brand-name counterpart (B). She has a researcher friend of hers run a
Small experiment to test $$\mathrm { H } _ { 0 } : \mu _ { G } - \mu _ { B } = 0 \text { against the alternative } \mathrm { H } _ { \mathrm { A } } : \mu _ { \mathrm { G } } - \mu _ { B } \neq 0 \text {. }$$
0) Which of the
Following would be a Type I error?
A) Deciding that the absorption rates are the same, when in fact they are not.
B) Deciding that the absorption rates are different, when in fact they are not.
C) The researcher cannot make a Type I error, since he has run an experiment.
D) Deciding that the absorption rates are different, when in fact they are.
E) Deciding that the absorption rates are the same, when in fact they are.

Accepted Answer

A Type I error occurs when the null hypothesis is rejected when it is actually true. In this context, it means deciding that the absorption rates are different when they are actually the same.

Question 10

Doctors at a technology research facility randomly assigned equal numbers of people to use computer keyboards in two rooms. In one room a group of people typed a manuscript using standard keyboards, while in the other room people typed the same manuscript using ergonomic keyboards to see if those people could type more words per minute. After collecting data for several days the researchers tested the hypothesis $\mathrm { H } _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ against the one-tail alternative and found $P = 0.22$. Which is true?
A) There's a 22% chance another experiment will give these same results.
B) none of these
C) The people using ergonomic keyboards type 22% more words per minute.
D) There's a 22% chance that there's really no difference in typing speed.
E) There's a 22% chance that people using ergonomic keyboards type more words per minute.

Accepted Answer

The p-value of 0.22 indicates that there is a 22% probability of observing the data, or something more extreme, assuming the null hypothesis is true. It does not provide the probability of the null hypothesis being true or the likelihood of future experiments yielding the same results.

Question 11

A researcher found that a 98% confidence interval for the mean hours per week spent studying by
College students was (13, 17). Which is true?
I. There is a 98% chance that the mean hours per week spent studying by college students is
Between 13 and 17 hours.
II. 98% of college students study between 13 and 17 hours a week.
III. Students average between 13 and 17 hours per week studying on 98% of the weeks.
A) III only
B) I and III
C) none of these
D) II only
E) I only

Accepted Answer

A confidence interval provides a range within which we are 98% confident the true population mean lies, not the probability of the mean being in that range. It also doesn't imply that 98% of students or weeks fall within that range.

Question 12

Food inspectors need to estimate the level of contaminants in food products packaged at a certain
Factory. Initial tests were based on a small sample but now the inspectors double the sample size for
A follow-up test. The main purpose of the larger sample is to…
A) reduce response bias.
B) decrease the standard deviation of the sampling model.
C) reduce non-response bias.
D) reduce confounding due to other variables.
E) decrease the variability in the population.

Accepted Answer

Doubling the sample size decreases the standard deviation of the sampling distribution, leading to more precise estimates of the population parameter.

Question 13

An elementary school principal wants to know the mean number of children in families whose
Children attend this school. He checks all the families using the school's registration records, and we
Create a 95% confidence interval based on a t-distribution. This procedure was not appropriate.
Why?
A) Since these families are from only one school, the family sizes may be skewed.
B) At a given school families are not randomly selected.
C) The entire population of families was gathered so there is no reason to do inference.
D) The recent record-setting family with twelve children is probably an outlier.
E) The population standard deviation is known, so he should have used a z-model.

Accepted Answer

The entire population of families was gathered so there is no reason to do inference.

Question 14

The two samples whose statistics are given in the table thought to come from populations with
Equal variances. What is the pooled estimate of the population standard deviation? $$\begin{array} { | c | c | c | } 
\hline n & \text { Mean } & \text { SD } \
\hline 50 & 22 & 3 \
\hline 55 & 25 & 4 \
\hline
\end{array}$$
A) 3.50
B) 1.87
C) 5.00
D) 3.56
E) 3.52

Accepted Answer

The answer of The two samples whose statistics are given...

Question 15

A contact lens wearer read that the producer of a new contact lens boasts that their lenses are
Cheaper than contact lenses from another popular company. She collected some data, then tested
The null hypothesis $$\mathrm { H } _ { 0 } : \mu _ { o l d } - \mu _ { n e w } = 0 \text { against the alternative } \mathrm { H } _ { \mathrm { A } } : \mu _ { o l d } - \mu _ { n e w } > 0$$
0) Which of the
Following would be a Type II error?
A) Deciding that the new lenses are cheaper, when in fact they really are.
B) Deciding that the new lenses are not really cheaper, when in fact they are not.
C) Deciding that the new lenses are cheaper, when in fact they are not.
D) Applying these results to all contact lenses, old and new.
E) Deciding that the new lenses are not really cheaper, when in fact they are.

Accepted Answer

A Type II error occurs when we fail to reject the null hypothesis when it is false. In this context, it means deciding that the new lenses are not really cheaper when, in fact, they are.

Question 16

A philosophy professor wants to find out whether the mean age of the men in his large lecture class is equal to the mean age of the women in his classes. After collecting data from a random sample of his students, the professor tested the hypothesis $\mathrm { H } _ { 0 } : \mu _ { M } - \mu _ { W } = 0$ against the alternative $\mathrm { H } _ { \mathrm { A } } : \mu _ { M } -$ $\mu W \neq 0$. The $P$-value for the test was $0.003$. Which is true?
A) There is a 99.7% chance that another sample will give these same results.
B) There is a 0.3% chance that the mean ages for the men and women are different.
C) There is a 0.3% chance that the mean ages for the men and women are equal.
D) It is very unlikely that the professor would see results like these if the mean age of men was equal to the mean age of women.
E) There is a 0.3% chance that another sample will give these same results.

Accepted Answer

The p-value of 0.003 indicates that if the null hypothesis were true (mean ages are equal), there is only a 0.3% chance of observing results as extreme as those obtained. This suggests it is very unlikely to see such results if the mean ages were actually equal.

Question 17

A contact lens wearer read that the producer of a new contact lens boasts that their lenses are
Cheaper than contact lenses from another popular company. She collected some data, then tested
The null hypothesis $$\mathrm { H } _ { 0 } : \mu _ { o l } d - \mu _ { n e w } = 0 \text { against the alternative } \mathrm { H } _ { \mathrm { A } } : \mu _ { o l } - \mu _ { n e w } > 0$$
0) Which of the
Following would be a Type II error?
A) Deciding that the new lenses are cheaper, when in fact they are not.
B) Deciding that the new lenses are not really cheaper, when in fact they are not.
C) Deciding that the new lenses are cheaper, when in fact they really are.
D) Applying these results to all contact lenses, old and new.
E) Deciding that the new lenses are not really cheaper, when in fact they are.

Accepted Answer

A Type II error occurs when the null hypothesis is not rejected when it is actually false. In this context, it means deciding that the new lenses are not really cheaper, when in fact they are.

Question 18

The two samples whose statistics are given in the table thought to come from populations with
Equal variances. What is the pooled estimate of the population standard deviation? $$\begin{array} { | c | c | c | } 
\hline n & \text { Mean } & \text { SD } \
\hline 50 & 22 & 3 \
\hline 55 & 25 & 4 \
\hline
\end{array}$$
A) 3.50
B) 3.56
C) 1.87
D) 5.00
E) 3.52

Accepted Answer

The pooled standard deviation is calculated using the formula: $\sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$. Substituting the given values: $\sqrt{\frac{(50-1)3^2 + (55-1)4^2}{50+55-2}} = \sqrt{\frac{49 \times 9 + 54 \times 16}{103}} = \sqrt{\frac{441 + 864}{103}} = \sqrt{\frac{1305}{103}} \approx 3.56$.

Question 19

Based on data from two very large independent samples, two students tested a hypothesis about
Equality of population means using $$\alpha = 0.05$$
5) One student used a one-tail test and rejected the null
Hypothesis, but the other used a two-tail test and failed to reject the null. Which of these might
Have been their calculated value of t?
A) 1.55
B) 1.88
C) 1.22
D) 2.22
E) 2.66

Accepted Answer

The answer of Based on data from two very large...

Question 20

A coffee house owner knows that customers pour different amounts of coffee into their cups. She
Samples cups from 10 costumers she believes to be representative of the customers and weighs the
Cups, finding a mean of 12.5 ounces and standard deviation of 0.5 ounces. Assuming these cups of
Coffee can be considered a random sample of all cups of coffee which of the following formulas
Gives a 95% confidence interval for the mean weight of all cups of coffee? 
A) $12.5 \pm 2.228 \left( \frac { 0.5 } { \sqrt { 10 } } \right)$
B) $12.5 \pm 2.262 \left( \frac { 0.5 } { \sqrt { 10 } } \right)$
C) $12.5 \pm 2.228 \left( \frac { 0.5 } { \sqrt { 9 } } \right)$
D) $12.5 \pm 2.262 \left( \frac { 0.5 } { \sqrt { 9 } } \right)$
E) $12.5 \pm 1.969 \left( \frac { 0.5 } { \sqrt { 10 } } \right)$

Accepted Answer

The correct formula for a 95% confidence interval for the mean when the sample size is small (n < 30) and the population standard deviation is unknown is $ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) $. Here, $ t^* $ is the t-score for 95% confidence with 9 degrees of freedom (n-1), which is approximately 2.262.

Quiz 6: Accessing Associations Between Variables

An elementary school principal wants to know the mean number of children in families whose Children attend this school. He checks all the families using the school's registration records, and we Create a 95% confidence interval based on a t-distribution. This procedure was not appropriate. Why?

A marketing company reviewing the length of television commercials monitored a random sample Of commercials over several days. They found that a 95% confidence interval for the mean length (in seconds) of commercials aired daily was (23, 27) . Which is true?

A professor was curious about her students' grade point averages (GPAs) . She took a random Sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of The following formulas gives a 99% confidence interval for the mean GPA of the professor's Students?

A survey asked people "On what percent of days do you get more than 30 minutes of vigorous Exercise?" Using their responses we want to estimate the difference in exercise frequency between Men and women. We should use a

We want to know the mean winning score at the US Open golf championship. An internet search Gives us all the scores for the history of that tournament, and we create a 95% confidence interval Based on a t-distribution. This procedure was not appropriate. Why?

Food inspectors need to estimate the level of contaminants in food products packaged at a certain Factory. Initial tests were based on a small sample but now the inspectors double the sample size for A follow-up test. The main purpose of the larger sample is to…

An elementary school principal wants to know the mean number of children in families whose Children attend this school. He checks all the families using the school's registration records, and we Create a 95% confidence interval based on a t-distribution. This procedure was not appropriate. Why?