. [16 points] A coach uses a nw technique to train gymnasts. 7 gymnasts were randomly selected and their competition scores were recorded before and after the training. The results are shown below.
Before After
9.7 9.9
9.4 9.7
9.6 9.2
9.6 9.4
9.6 9.7
9.7 10.0
9.4 9.2
Using a 0.05 level of significance, test the claim that the training technique is effective in raising the gymnasts’ scores.
Ho:
Ha:
Test Statistic:
p-value:
Decision:
Conclusion:
2.[12 pts] A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to construct a 90% confidence interval for the difference in blood pressure. Can you conclude that the diet helped? (Assume the two samples are independent and have been randomly selected.)
Treatment Group Control Group
n=85 n=51
mean=189.1 mean=203.7
s=38.7 s=37.2
3.[16 pts] A researcher finds that of 1000 people who said that they attend a religious service at least once a week, 30 stopped to help a person with car trouble. Of 1200 people interviewed who had not attended a religious service at least once a month, 20 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are equal. (Assume the samples are independent and have been randomly selected.)
Ho:
Ha:
Test Statistic:
Pvalue:
Decision:
Conclusion:
4.[14 pts] Managers rate employees according to job performance and attitude. The results for several randomly selected employees are shown below.
Performance Attitude
____________ ______________
59 72
63 67
65 78
69 82
58 75
77 87
76 92
69 83
70 87
64 78
a). Use a 0.05 significance level to test for a linear correlation between attitude and performance.
Ho:
Ha:
Test Statistic:
Critical Value(s):
Decision:
Conclusion:
b) What percentage of the variation in performance can be explained by the linear relationship?
c) Find the equation of the regression line that expresses performance (y) in terms of attitude (x).
d) What is the best predicted performance rating for a person whose attitude rating is 80?
5.[16 pts] In studying the responses to a multiple-choice test question, the following sample data were obtained. At the .05 significance level, test the claim that the responses occur with the same frequency.
(Show all calculations necessary to calculate the test statistic.)
Response Frequency
A 11
B 15
C 17
D 18
E 19
Ho:
Ha:
Test Statistic:
Pvalue:
Decision:
Conclusion:
6. [16 pts] A case-control study was conducted to investigate a relationship between colors of helmets worn by motorcycle drivers and whether they are injured or killed in a crash. Results are summarized below. Test the claim that injuries are independent of helmet color, using a .05 level of significance. Should motorcycle drivers chose helmets with a particular color? If so, which color appears best?
Black White Yellow/Orange Red Blue
Controls (not injured) 481 394 30 167 58
Cases (injured or killed) 198 120 5 73 27
7. [10 pts] Below are data from the 1st and 2nd exams in this class. Find the best model. Consider only linear, quadratic, logarithmic, exponential, and power models. Indicate r2 for each model.
Exam 1(X) Exam 2(Y)
69 92.5
87 90
51 71.5
43.5 30
93 93
91 71.5
96 97
98.5 96.5
84 96.5
97 93.5
73 60
82.5 56
83 83.5
85 73.5
93 79
96 97
40 30.5
82 81
89 73
64.5 31
90 97
35 29
83 77
82 38
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1. Hypothesis Test
Null Hypothesis (Ho): The training technique is not effective in raising the gymnasts’ scores.
Alternative Hypothesis (Ha): The training technique is effective in raising the gymnasts’ scores.
To test this claim, we can perform a paired t-test since we have before and after scores for the same gymnasts.
Test Statistic: t = (mean(After) - mean(Before)) / (s / sqrt(n))
where mean(After) and mean(Before) are the mean scores after and before the training, respectively.
s is the standard deviation of the differences in scores.
n is the number of gymnasts.
p-value: The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true.
Decision: We compare the p-value to the significance level (0.05) and make a decision.
If the p-value is less than 0.05, we reject the null hypothesis.
If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
Conclusion: We interpret the results of the test and make a conclusion based on our decision.
2. Confidence Interval
To construct a 90% confidence interval for the difference in blood pressure between the treatment and control groups, we can use the formula:
Confidence Interval: mean(Treatment) - mean(Control) ± t * (s / sqrt(n)),
where mean(Treatment) and mean(Control) are the means of the treatment and control groups, respectively.
s is the pooled standard deviation.
n is the number of individuals in each group.
t is the critical value corresponding to a 90% confidence level.
We can conclude that the diet helped if the confidence interval does not include zero (no effect).
3. Hypothesis Test
Null Hypothesis (Ho): The proportions of people who stopped to help with car trouble are the same for those who attend a religious service at least once a week and those who do not attend at least once a month.
Alternative Hypothesis (Ha): The proportions of people who stopped to help with car trouble are different for those who attend a religious service at least once a week and those who do not attend at least once a month.
To test this claim, we can perform a two-sample proportion test.
Test Statistic: z = (p1 - p2) / sqrt(p * (1 - p) * (1/n1 + 1/n2))
where p1 and p2 are the sample proportions.
p is the pooled proportion.
n1 and n2 are the sample sizes.
p-value: The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true.
Decision: We compare the p-value to the significance level (0.05) and make a decision.
If the p-value is less than 0.05, we reject the null hypothesis.
If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
Conclusion: We interpret the results of the test and make a conclusion based on our decision.
4. Correlation Test
a) Hypothesis Test
Null Hypothesis (Ho): There is no linear correlation between attitude and performance.
Alternative Hypothesis (Ha): There is a linear correlation between attitude and performance.
We can test for linear correlation using Pearson's correlation coefficient.
Test Statistic: r = (n * Σ(xy) - Σx * Σy) / sqrt((n * Σx^2 - (Σx)^2) * (n * Σy^2 - (Σy)^2))
where n is the number of data points,
Σxy is the sum of the products of corresponding values of the two variables,
Σx and Σy are the sums of the x and y values, respectively,
Σx^2 and Σy^2 are the sums of the squared x and y values, respectively.
Critical Value: To determine the critical value, we need to specify the significance level and degrees of freedom. For example, at a 0.05 significance level with (n-2) degrees of freedom, we can look up the critical value for correlation coefficient from a t-table.
Decision: We compare the test statistic to the critical value and make a decision.
If the test statistic is greater than the critical value, we reject the null hypothesis.
If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.
Conclusion: We interpret the results of the test and make a conclusion based on our decision.
b) Explained Variation
The percentage of the variation in performance explained by the linear relationship can be determined by squaring the correlation coefficient (r) and multiplying by 100.
Explained Variation = (r^2) * 100
c) Regression Line
The equation of the regression line that expresses performance (y) in terms of attitude (x) can be determined from the linear regression analysis. The equation has the form:
y = a + bx
where a is the y-intercept,
b is the slope of the line.
d) Predicted Performance
To find the best predicted performance rating for a person whose attitude rating is 80, we substitute x = 80 into the regression equation and solve for y.
5. Hypothesis Test
Null Hypothesis (Ho): The responses occur with the same frequency.
Alternative Hypothesis (Ha): The responses do not occur with the same frequency.
We can perform a chi-square test for goodness of fit to test this claim.
Test Statistic: χ^2 = Σ((O - E)^2 / E),
where O is the observed frequency,
E is the expected frequency.
p-value: The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true.
Decision: We compare the p-value to the significance level (0.05) and make a decision.
If the p-value is less than 0.05, we reject the null hypothesis.
If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
Conclusion: We interpret the results of the test and make a conclusion based on our decision.
6. Independence Test
To test the claim that injuries are independent of helmet color, we can perform a chi-square test for independence.
Test Statistic: χ^2 = Σ((O - E)^2 / E),
where O is the observed frequency,
E is the expected frequency.
p-value: The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true.
Decision: We compare the p-value to the significance level (0.05) and make a decision.
If the p-value is less than 0.05, we reject the null hypothesis.
If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
Conclusion: We interpret the results of the test and make a conclusion based on our decision.
7. Model Selection
We can fit different models (linear, quadratic, logarithmic, exponential, power) to the data and calculate the r^2 (coefficient of determination) for each model. The model with the highest r^2 would be considered the best model.