MATHEMATICS

12.Interpret the given confidence interval.

A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Subjects were randomly assigned to either a treatment group or a control group. The mean blood pressure was determined for each group, and a 95% confidence interval for the difference in the means for the treatment group versus the control group, Ut-Uc was found to be (-21, -6). ( t –stands for treatment group and C stands for control group)

  1. We are 95% confident that the average blood pressure of those who follow the diet is between 6 and 21 points higher than the average for those who do not follow the diet.
  2. The probability that the mean blood pressure for those on the diet is lower than for those not on the diet is 0.95.
  3. Since all of the values in the confidence interval are less than 0, we are unable to conclude that there is difference in blood pressure for those who follow the diet and those who do not.
  4. We are 95% confident that the average blood pressure of those who follow the diet is between 6 and 21 points lower than the average for those who do not follow the diet.
  5. The probability that the mean blood pressure for those on the diet is higher than for those not on the diet is 0.95.

13.Interpret the given confidence interval.

A researcher was interested in comparing the salaries of female and male employees of a particular company. Independent random samples of female employees (sample 1) and male employees (sample 2) were taken to calculate the mean salary, in dollars per week, for each group. A 90% confidence interval for the difference, U1-U2 between the mean weekly salary of all female employees and the mean weekly salary of all male employees was determined to be (-$110,$10)

  1. 90% of the time females at this company make less than males
  2. Since 0 is contained in the interval, the probability that the male employees at this company earn the same as females at this company is 0.9
  3. Based on these data, we are 90% confident that the male employees at this company average between $110 less and $10 more per week than the female employees.
  4. The probability that a randomly selected female employee at this company makes between $110 less and $10 more per week than a randomly selected male employee is 0.9.
  5. Based on these data, we are 90% confident that the female employees at this company average between $ 110 less and $10 more per week than the male employees.

14. In a positive association between two variables

  1. A decrease in the value of one variable is associated with a decrease in the value of a second variable.
  2. An increase in the value of one variable is associated with a decrease in the value of a second variable.
  3. An increase in the value of one variable is associated with an increase in the value of a second variable.
  4. Both A and C

15.Comparison of means can be used when

  1. A researcher wants to compare responses to an ordinal variable by the categories of a nominal variable.
  2. A researcher wants to compare responses to a numerical variable by the categories of a nominal or ordinal variable
  3. A researcher wants to compare responses to a nominal variable by the categories of an ordinal variable
  4. A researcher wants to compare responses to a numerical variable by categories of an ordinal variable

16. The independent samples t test assesses differences in means between

  1. An independent nominal variable and dependent ordinal variable
  2. A dependent numerical variable and an independent categorical variable
  3. A dependent nominal variable and an independent categorical variable
  4. A dependent categorical variable and an independent numerical variable

 

17. In a negative association between two variables,

  1.  A decrease in the value of one variable is associated with a decrease in the value of a second variable
  2. An increase in the value of one variable is associated with a decrease in the value of a second variable
  3. A decrease in the value of one variable is associated with an increase in the value of a second variable
  4. Both B and C

18.For the independent samples t test, if equality of variances cannot be assumed then you

  1. Have to use a formula for the t statistic that is different from the one you would use if equality of variances can be assumed
  2. Have to use a formula for degrees of freedom that is different from the one you would use if equality of variances can be assumed
  3. Cannot reject the null hypothesis , no matter the value of t
  4. Both A and B

19. Provide an appropriate response.

A 95% confidence interval for the difference in means for a collection of paired sample data is (0, 3.4) Based on the same sample, a traditional significance test fails to support the claim of μd > 0. What can you conclude about the significance level α (α = 1 – .95) of the hypothesis test?

 

α > 0.05

α < 0.05

α = 0.01

α = 0.05

α = 0.95

20.From the sample statistics, find the value of P1cap-P2cap, the point estimate of the difference of proportions. Unless otherwise indicated, round to the nearest thousandth when necessary.

n1 = 100  n2 = 100

x1 = 34  x2 =  30

 

-0.04

none of these

-0.02

0.02

0.04

20. The statistic that answers the question, how likely is it that the difference between the means for two categories of a variable that we observe in a sample is merely a chance occurrence, is the

 

independent samples t-test

t statistic for Pearson’s r

one-sample t test

one-way analysis of variance

21.Interpret the given confidence interval.

A high school coach uses a new technique in training middle distance runners. He records the times for 4 different athletes to run 800 meters before and after this training. A 90% confidence interval for the difference of the means before and after the training, μB – μA, was determined to be(2.7, 4.2)

 

The probability that the average time for the 800-meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds shorter after the training is 0.9.

We are 90% confident that a randomly selected middle distance runner at this high school will have a time for the 800-meter run that is between 2.7 and 4.2 seconds shorter after the training than before the training.

Based on this sample, we are 90% confident that the average time for the 800-meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds longer after the new training.

Based on this sample, we are 90% confident that the average time for the 800-meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds shorter after the new training.

We know that 90% of the middle distance runners shortened their times between 2.7 and 4.2 seconds after the training.

22.  Select the most appropriate answer.

The central limit theorem predicts that the sampling distribution of X1bar – X2bar is approximately normal

 

when the total number sampled is greater than or equal to 30.

when both of the sample sizes are greater than or equal to 30.

when either one of the sample sizes is greater than or equal to 30.

when at least one of the sample sizes is greater than or equal to 30.

regardless of both of the sample sizes.

From the sample statistics, find the value of P1cap-P2cap , the point estimate of the difference of proportions. Unless otherwise indicated, round to the nearest thousandth when necessary.

A survey asked respondents whether marijuana should be made legal. A 95% confidence interval for PA-PB, is given by (0.08,0.14) where PA is the proportion of respondents who answered “legal” in state A and PB is the proportion of respondents who responded “legal” in state B. Based on the 95% confidence interval, what can we conclude about the percentage of respondents who favor legalization in state B versus state A?

 

Since all of the values in the confidence interval are less than 1, we can conclude that there is a significant difference between the percentage in favor of legalization in state B and the percentage in favor of legalization in state A.

Since all of the values in the confidence interval are less than 1, we are unable to conclude that there is a significant difference between the percentage in favor of legalization in state B and the percentage in favor of legalization in state A.

Since all of the values in the confidence interval are greater than 0, we can conclude that the percentage in favor of legalization was greater in state A than it was in state B.

Since all of the values in the confidence interval are greater than 0, we can conclude that the percentage in favor of legalization was greater in state B than it was in state A.

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