Suppose a Mathematics lecturer believes that there is no significant difference between the average class scores of his two classes, BNR and BNF on Final Exam. The average and standard deviation for Class BNR of 35 students were 75.86 and 16.91, respectively. The average and standard deviation for Class BNF of 37 students were 75.41 and 19.73, respectively. Using these results, test the claims at 5% level of significance.
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Q: You are testing the claim that the mean GPA of night students is different from the mean GPA of day…
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Q: You are testing the claim that the mean GPA of night students is different from the mean GPA of day…
A: given data NIGHT STUDENT :x¯1 = 2.44s1 =0.33n1 = 35MBAx¯2 =2.21s2 = 0.45n2 = 60α = 0.05claim : μ1…
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A: Given: n1=30x¯1=2.83s1=0.94n2=55x¯2=2.97s2=0.92α=0.01
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Q: You are testing the claim that the mean GPA of night students is different from the mean GPA of day…
A: given data GPA night x¯1 = 2.63s1 = 0.81n1 = 45GPA of day x¯2 = 3.02s2 = 0.59n2 = 40α = 0.05claim…
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A: Given For males, μ=4.296σ=0.907 For females μ=3.764σ=0.866
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Q: A professor believes that, for the introductory art history classes at his university, the mean test…
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Q: You are testing the claim that the mean GPA of night students is different from the mean GPA of day…
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Q: A professor believes that, for the introductory art history classes at his university, the mean test…
A: The objective of this question is to test the professor's claim that the mean test score of students…
Q: One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings…
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Q: One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings…
A: Answer Given Terry:Mean =4.364Standard deviation =0.639Karen:Mean =3.801Standard deviation =0.959
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- A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2For all students who have taken a certain class during the last ten years, theaverage on the mid-term is 83.4 and the standard deviation of the mid-termscores is 7.2; the average on the final is 79.3 and the standard deviation of thescores on the final is 5.1; the correlation between scores on the mid-term andscores on the final is 0.86. 4. My teacher has lost our mid-term scores (but at least has our scores onthe final), so she is fitting a straight line to guess what we probably got onthe mid-term. What’s the equation of the line she is using? 5. I got a 90 on the final. What does my teacher think I got on the midterm?You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 45 night students, and the sample mean GPA is 2.63 with a standard deviation of 0.81. You sample 40 day students, and the sample mean GPA is 3.02 with a standard deviation of 0.59. Test the claim using a 5% level of significance. Assume the population standard deviations are unequal and that GPAS are normally distributed. Give answer to at least 4 decimal places. What are the correct hypotheses? Ho: M₁ = μ₂ H₁: M₁ Based on the hypotheses, find the following: Test Statistic = -0.7918 X O μ₂ Critical Values=2.0154 x (Just enter the positive CV.) The correct decision is to Reject the null hypothesis The correct summary would be: There is enough evidence to support the claim of night students is different from the mean GPA of day students. o that the mean GPA
- Use z scores to compare the given values. The tallest living man at one time had a height of 238 cm. The shortest living man at that time had a height of 142.4 cm. Heights of men at that time had a mean of 175.45 cm and a standard deviation of 5.59 cm. Which of these two men had the height that was more extreme? ... Since the z score for the tallest man is z = 0 and the z score for the shortest man is z = the man had the height that was Im- more extreme. (Round to two decimal places.) shortest tallestHow could I solve these problems?You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 35 night students, and the sample mean GPA is 2.91 with a standard deviation of 0.97. You sample 20 day students, and the sample mean GPA is 2.99 with a standard deviation of 0.82. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAS are normally distributed. Give answer to at least 4 decimal places. What are the correct hypotheses? Ho: M₁ H₁: M₁ Test Statistic = Critical Values = = ± H₁₂ Based on the hypotheses, find the following: ≠v H₂ OF (Just enter the positive CV.) The correct decision is to Fail to reject the null hypothesis The correct summary would be: There is not enough evidence to support the claim ✓ of night students is different from the mean GPA of day students. o that the mean GPA
- You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.83 with a standard deviation of 0.94. You sample 55 day students, and the sample mean GPA is 2.97 with a standard deviation of 0.92. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAs are normally distributed. Give answer to at least 4 decimal places. What are the correct hypotheses?H0: Select an answer μ σ² μ₂ p μ₁ x̄₂ s² x̄₁ = Select an answer μ x̄₂ σ² μ₂ x̄₁ μ₁ s² p H1: Select an answer σ² μ x̄₁ s² p x̄₂ μ₁ μ₂ ? = < > ≤ ≠ ≥ Select an answer μ x̄₂ x̄₁ μ₂ σ² s² μ₁ p Based on the hypotheses, find the following:Test Statistic = Critical Values = ±± (Just enter the positive CV.)The correct decision is to Select an answer Reject the null hypothesis Fail to reject the null hypothesis The correct summary would be: Select an answer There is not enough…2. Alfred scored 82 in a quiz in English for which the average Score of the class was 75 with a standard deviation of 10. He also took a quiz in Statistics and scored 70 for which the average score of the class was 55, and the standard deviation was 14. Relative to other students in the class, did Alfred do better in English or Statistics?One year Hank had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 3.09. Also, Karen had the lowest ERA of any female pitcher at the school with an ERA of 2.98. For the males, the mean ERA was 4.635 and the standard deviation was 0.576. For the females, the mean ERA was 4.469 and the standard deviation was 0.944. Find their respective z-scores. Which player had the better year relative to their peers, Hank or Karen? (Note: In general, the lower the ERA, the better the pitcher.) Hank had an ERA with a z-score of Karen had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers?
- A recent study found that 61 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 39 grams of potato chips as compared to a mean of 28 grams for 51 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 21.2 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.7 grams. What is the test statistic? tSTAT= __________ (Round to two decimal places as needed.)A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 85.6. He knows the population standard deviation for the evening classes to be 4.6 points. A random sample of 150 students from morning classes results in a mean test score of 86.7. He knows the population standard deviation for the morning classes to be 8.3 points. Test his claim with a 99 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3: Draw a conclusion and interpret the decision. Answer 围 Tables E Keypad Keyboard Shortcuts We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the…A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 86.6. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 87.8. He knows the population standard deviation for the morning classes to be 4.7 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.