1. On average, a doctor will treat between 0 and 5 patients a day. The random variable x represents the number of patients treated on a given day.
x 0 1 2 3 4 5
P(x) 0.01 0.11 0.35 0.42 0.05 0.06
This probability distribution results in a mean score of approximately 2.57. How do you interpret the mean?
a. The doctor will treat an average of 3 patients each day
b. The doctor will treat an average of 5 patients each day
c. The number of patients treated differs
d. The number of patients that the doctor will treat varies by 3 patients
2. The Director of the Medical Board is interested in determining if it is “unusually low” for a doctor to treat one patient each day. The Director determines that the probability of a doctor treating 1 patient or fewer patients is 0.12. Based on the rules of using probabilities to determine when results are unusual, what should the director conclude?
a. It would not be unusually low because the probability of 0 patients treated is smaller than the probability of 1 patient being treated.
b. It would be unusually low because the probability of 0.12 is small.
c. It would not be unusually low because the probability of 0.12 is greater than the 0.05 requirement.
d. It would be unusually low because the probability of 0.12 is significantly less than the average.
3. Considering a standard normal distribution, find the area under the standard normal curve between z = -1.5 and z = 2.25
4. A car manufacturer is interested in determining the average braking distance for select tires under wintery conditions. He selects a random sample of select tires and assumes that the average braking distances are normally distributed. Which of the following would allow the manufacturer to reject the normality assumption?
a. There were no outliers in the sample
b. The shape of the histogram shows a distribution that was “bell-shaped.”
c. The quantile plot shows points that are not in a straight-line pattern.
d. The quantile plot shows points that are lying in a straight-line pattern.
5. Twenty randomly selected students took the statistics midterm exam. The results from the midterm show that we have a sample mean of 75 and a margin of error of 4.28. Construct the confidence interval for the mean score of all students.
a. 70.72 < <79.28
b. 75.00 < <79.28
c. 70.72 < <75.00
d. 72.86 < <77.14
6. A 95% confidence interval for the population proportion of adults that use a cell phone versus a land-line phone for their primary means of communications is 73.2% < p < 82.4%. Which answer choice best interprets the results of this confidence interval?
a. We are 5% confident that the true population of adults that use a cell phone as their primary means of communication is between 73.2% and 82.4%.
b. We are 5% confident that the true population of adults that use a cell phone as their primary means of communication is greater than 82.4%.
c. We are 95% confident that the true population of adults that use a cell phone as their primary means of communication is between 73.2% and 82.4%.
d. We are 95% confident that the true population of adults that use a cell phone as their primary means of communication is less than 73.2%.
7. The P-value for a hypothesis test is P = 0.0325. What is your decision if your significance level is = 0.05.
a. 0.0325 < 0.05; you should reject the null hypothesis.
b. 0.0325 > 0.05; you should reject the null hypothesis.
c. 0.0325 < 0.05; you should fail to reject the hypothesis.
d. 0.0325 > 0.05; you should fail to reject the hypothesis.
8. A restaurant owner wants to determine if there is enough evidence that the proportion of wait times to be seated in a restaurant is higher for group A than group B. If the samples that were obtained for each group are randomly selected and independent of one another, which test should the restaurant owner use to test the difference in wait times for group A and group B?
c. X2 (chi-test)
d. None of the above
9. The table below shows a one-way analysis of variance with sample data that consists of measuring 5th grade math scores from 12 randomly selected pages in 3 different textbooks.
Source of variation SS df MS F P-value F-crit
Between Groups 68.19 2 34.09 ??? 0.001 3.2849
Within groups 125.31 33 3.80
Total 193.50 35
Based on the results of ANOVA, determine the F-statistic.
Answer should be F = 34.09/3.80 = 8.97 but it’s not one of the option.
10. Using the results from the ANOVA table above, what is the null hypothesis and what is the final conclusion?
a. Ho: 1 = 2 = 3; there is not sufficient evidence to reject the null hypothesis that the 5th grade math scores for all three populations have the same mean.
b. Ho: 1 = 2 = 3; there is sufficient evidence to reject the null hypothesis that the 5th grade math scores for all three populations have the same mean.
c. Ho: 1 > 2 > 3; there is sufficient evidence to support the null hypothesis that the 5th grade math scores for all three populations have the same mean.
d. Ho: 1 < 2 < 3; there is not sufficient evidence to support the null hypothesis that the 5th grade math scores for all three populations have the same mean.
11. A teacher wants to test whether the hours that a student spends studying online (independent variable) is correlated with the students test scores (dependent variable). To determine if the variables were correlated with the students test scores, the instructor calculated the correlation coefficient, r, of -0.831. What conclusion can the instructor make about the correlation of these two variables?
a. There is a weak positive correlation.
b. There is a strong positive correlation.
c. There is a weak negative correlation.
d. There is a strong negative correlation.
12. A study was done to determine whether shoe size and height correlated. It was found that there is a significant correlation between the two variables. Given that the equation of the regression line is ŷ = 1.87 + 51.36, where x = shoe size and ŷ = height in inches, what height would you predict for someone with a shoe size of 10?
a. ŷ = 70.06
b. ŷ = 72.87
c. ŷ = 66.32
d. ŷ = 69.57
13. A study shows that the number of crimes committed in a randomly selected city is uniformly distributed. The researcher randomly selected a city and randomly selected 2000 crimes from a recent year and conducted a goodness of fit test. The null hypothesis is the researcher’s claim that the distribution of the number of crimes is uniform. At = 0.10, the test produced a X2 (chi squared) critical value of 17.275, a rejection region X2 > 17.275 and X2 test statistic of 10.32. What can you conclude from this goodness of fit test?
a. Fail to reject the Ho; there is not enough evidence at the 10% level of significance to reject the researcher’s claim.
b. Fail to reject the Ha; there is enough evidence at the 10% level of significance to reject the researcher’s claim.
c. Reject the Ho
d. Reject the Ha
14. Below is a contingency table that contains the frequencies of 125 randomly selected criminal cases. The row variable identifies the conclusion of the trial (conclusion or no conviction) and the column variable identifies the plea (guilty or not guilty). Based on the table below, determine the percentage of criminal cases that ended in convictions and the plea was guilty.
Guilty Not guilty
Conclusion 36 20
a. About 45%
b. About 29%
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