Question 1

Exhibit 7-4: A group of researchers has developed a method that they hypothesize will improve physics test scores.They randomly selected nine students to participate in their study.The nine students were administered an ACT test, trained in the new method, and then administered an alternate form of the ACT test.The following data are their scores:
 $ \begin{array}{lll}\text {Subject } & {\text {Test 1 }}{} & \text {Test 2 } \ 1 & 20 & 19 \ 2 & 20 & 20 \ 3 & 24 & 21 \ 4 & 24 & 24 \ 5 & 24 & 22 \ 6 & 17 & 15 \ 7 & 18 & 15 \ 8 & 18 & 16 \ 9 & 20&  18\end{array} $
-Refer to Exhibit 7-4.What is the null hypothesis?

Accepted Answer

The answer of Exhibit 7-4: A group of researchers has...

Question 2

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.What is the null hypothesis?

Accepted Answer

The null hypothesis is typically stated as there being no significant difference between the two groups being compared. Therefore, the null hypothesis in this case is that the mean algebra test score for the computer-based training group (CBT) is equal to the mean algebra test score for the control group. This can be written as:  $\mu$CBT =  $\mu$Control.

Question 3

Exhibit 7-4: A group of researchers has developed a method that they hypothesize will improve physics test scores.They randomly selected nine students to participate in their study.The nine students were administered an ACT test, trained in the new method, and then administered an alternate form of the ACT test.The following data are their scores:
 $ \begin{array}{lll}\text {Subject } & {\text {Test 1 }}{} & \text {Test 2 } \ 1 & 20 & 19 \ 2 & 20 & 20 \ 3 & 24 & 21 \ 4 & 24 & 24 \ 5 & 24 & 22 \ 6 & 17 & 15 \ 7 & 18 & 15 \ 8 & 18 & 16 \ 9 & 20&  18\end{array} $
-Refer to Exhibit 7-4.The test is statistically significant at what level?

Accepted Answer

To determine the level of significance, we need to conduct a paired t-test. Using a calculator or software, we find that the t-value is -2.89 with 8 degrees of freedom. Looking at a t-distribution table, we find that the critical t-value at .01 significance level is 2.896. Since our calculated t-value (-2.89) is less than the critical t-value (-2.896), we can reject the null hypothesis at the .01 significance level. Therefore, the test is statistically significant at the .01 level, but not at the .05 or .001 levels.

Question 4

Exhibit 7-4: A group of researchers has developed a method that they hypothesize will improve physics test scores.They randomly selected nine students to participate in their study.The nine students were administered an ACT test, trained in the new method, and then administered an alternate form of the ACT test.The following data are their scores:
 $ \begin{array}{lll}\text {Subject } & {\text {Test 1 }}{} & \text {Test 2 } \ 1 & 20 & 19 \ 2 & 20 & 20 \ 3 & 24 & 21 \ 4 & 24 & 24 \ 5 & 24 & 22 \ 6 & 17 & 15 \ 7 & 18 & 15 \ 8 & 18 & 16 \ 9 & 20&  18\end{array} $
-Refer to Exhibit 7-4.What is the observed value for the chosen statistic?

Accepted Answer

The observed value for the chosen statistic is calculated using the differences between Test 1 and Test 2 scores. The differences are: 1, 0, 3, 0, 2, 2, 3, 2, 2. The mean of these differences is 1.6667, and the standard deviation is approximately 1.118. Using these, the t-statistic is calculated as 4.603.

Question 5

A school administrator finds no statistically significant difference in learning whether students are taught by computer or by books.She decides not to spend money on new computer equipment.Later, she finds that a neighboring district is very successful in implementing computer technology.

Accepted Answer

By deciding not to spend money on new computer equipment based on the assumption that there was no statistically significant difference in learning between students taught by computers or books, the school administrator may have made a Type II error or a false negative. This means that she wrongly concluded that there was no difference in learning between the two methods of teaching, when there actually was a difference. The successful implementation of computer technology in a neighboring district further supports the possibility of the school administrator making a Type II error.

Question 6

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.The researcher can conclude that:

Accepted Answer

To determine if there is a statistically significant difference between the two groups, a hypothesis test needs to be conducted. This requires calculating the mean and standard deviation for each group, and then using a t-test to compare the means. However, just from looking at the scores, it is clear that there is a range of scores in each group, with some overlap. Therefore, it is not accurate to choose B, as there are some students who scored lower with the computer program than those who did not use it. So, the best choice is C, as there is evidence that the means of the two groups are different.

Question 7

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.The test is statistically significant at what level (two-tailed)?

Accepted Answer

The answer of Exhibit 7-3: A researcher is examining the...

Question 8

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.The degrees of freedom for the test are the

Accepted Answer

The degrees of freedom for an independent samples t-test is calculated as the total number of cases minus 2.

Question 9

If a research study concludes that a new computer-based reading program teaches students how to read more quickly than the old method

Accepted Answer

A Type I error occurs when a study incorrectly rejects the null hypothesis, suggesting that the new program is effective when it might not be.

Question 10

"Do athletes who include imaging their success as part of their practice do better than athletes who do not include imaging?" The null hypothesis is

Accepted Answer

The null hypothesis always assumes that there is no difference or no effect between two groups, in this case between athletes who include imaging in their practice and athletes who do not include imaging. Therefore, the correct null hypothesis is C, which states that there is no difference in performance between the two groups.

Question 11

Exhibit 7-4: A group of researchers has developed a method that they hypothesize will improve physics test scores.They randomly selected nine students to participate in their study.The nine students were administered an ACT test, trained in the new method, and then administered an alternate form of the ACT test.The following data are their scores:
 $ \begin{array}{lll}\text {Subject } & {\text {Test 1 }}{} & \text {Test 2 } \ 1 & 20 & 19 \ 2 & 20 & 20 \ 3 & 24 & 21 \ 4 & 24 & 24 \ 5 & 24 & 22 \ 6 & 17 & 15 \ 7 & 18 & 15 \ 8 & 18 & 16 \ 9 & 20&  18\end{array} $
-Refer to Exhibit 7-4.What test is most appropriate for this data and hypothesis?

Accepted Answer

A correlated t-test is the most appropriate test for this data, as it involves comparing the scores of the same set of individuals before and after an intervention (in this case, before and after being trained in the new method). The other options (chi square, independent t-test, Pearson's r) are not appropriate for this type of analysis.

Question 12

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.What is the observed value for the chosen statistic?

Accepted Answer

To find the observed value for the chosen statistic, we need to calculate the t-statistic. Using a calculator or software, we can find that the mean score for the CBT group is 52.14 and the mean score for the control group is 47.29. The sample standard deviation for the CBT group is 2.20 and for the control group is 12.55. Using these values, we can calculate the t-statistic as follows:

t = (52.14 - 47.29) / sqrt((2.20^2 / 8) + (12.55^2 / 8))
t = 2.85

Therefore, the observed value for the chosen statistic is 2.846, which corresponds to choice B.

Question 13

A researcher wants a 90 percent chance of rejecting the null hypothesis at the one-tailed .05 level with an effect size of .15.The sample size needed is about

Accepted Answer

To calculate the sample size needed for this study, we can use a power analysis. Given a target power of .90, a one-tailed alpha level of .05, and an effect size of .15 (which is considered a small effect size), we can use a power analysis calculator or software to determine the necessary sample size.

Using an online calculator (such as G*Power), we find that the sample size needed is approximately 380 participants. This means that if we recruit 380 participants and run our study, we will have a 90% chance of detecting a statistically significant effect (i.e. rejecting the null hypothesis) at the .05 level, assuming the effect size is true.

Therefore, the best choice is E.

Question 14

Which one of the following statements is true?

Accepted Answer

The relative seriousness of a Type I or a Type II error is a judgment, as it depends on the specific context and consequences of each error. For example, in medical diagnosis, a Type II error (failing to diagnose a serious illness) may have more severe consequences than a Type I error (diagnosing an illness when there is none). Therefore, there is no inherent superiority of one error over the other.  The other statements are not accurate.

Question 15

The greatest statistical power occurs when

Accepted Answer

The greatest statistical power occurs when the sample size is large and the heterogeneity is small. A large sample size increases the precision of the estimate and reduces the impact of random sampling error, while low heterogeneity reduces the variability in the sample and improves the internal validity of the study. This increases the likelihood of detecting a true effect if one exists, making the statistical power of the study higher.

Question 16

An administrator will implement a new science curriculum if there is good evidence that it is more effective than the present curriculum.Otherwise he will continue with the present curriculum.The administrator should use a

Accepted Answer

A directional test is appropriate because the administrator is specifically interested in whether the new curriculum is more effective than the current one, indicating a specific direction (better, not just different) of the expected outcome.

Question 17

Exhibit 7-3: A researcher is examining the effects of a computer-based training program designed to teach algebra.The researcher randomly selects subjects for two groups and gives one group the computer training and the other standard teaching methods to see if the results of the two methods differ.The following scores are from the subjects' final test.
 $ \begin{array}{ll}{\text {CBT }}{} & \text {Control }\ 52 & 65  \ 55 & 45 \ 56 & 47 \ 53 & 46 \ 50 & 47 \ 49 & 33 \ 52 & 39 \ 53 & 49\end{array} $
-Refer to Exhibit 7-3.Which test is most appropriate for these data and hypothesis?

Accepted Answer

The most appropriate test for comparing two independent groups of scores is the independent t-test. Chi square is used for examining categorical data, while Pearson's r measures the strength and direction of a linear relationship between two continuous variables. Correlated t-test is used when there are two related samples (e.g., pre-test and post-test scores).

Question 18

If a test has been found to be statistically significant at the .05 level, the probability of getting this result by chance alone is

Accepted Answer

A statistically significant result at the .05 level means that the probability of getting a result as extreme as the one observed by chance alone is equal to or less than 5 times out of 100. Therefore, the correct choice is C.

Question 19

The null hypothesis states that

Accepted Answer

The null hypothesis is the assumption that there is no significant difference or relationship between two variables. It states that any observed difference or relationship is due to chance alone.

Question 20

Exhibit 7-4: A group of researchers has developed a method that they hypothesize will improve physics test scores.They randomly selected nine students to participate in their study.The nine students were administered an ACT test, trained in the new method, and then administered an alternate form of the ACT test.The following data are their scores:
 $ \begin{array}{lll}\text {Subject } & {\text {Test 1 }}{} & \text {Test 2 } \ 1 & 20 & 19 \ 2 & 20 & 20 \ 3 & 24 & 21 \ 4 & 24 & 24 \ 5 & 24 & 22 \ 6 & 17 & 15 \ 7 & 18 & 15 \ 8 & 18 & 16 \ 9 & 20&  18\end{array} $
-Refer to Exhibit 7-4.The degrees of freedom for the test are the

Accepted Answer

The degrees of freedom for a paired samples t-test, which is appropriate for this scenario where the same subjects are measured before and after an intervention, is calculated as the number of pairs (subjects) minus one. Here, there are 9 pairs (subjects), so the degrees of freedom are 9 - 1 = 8.

Quiz 7: Sampling and Inferential Statistics