An educational psychologist is studying two methods of tutoring young children: Method A and Method B. She wants to test whether there is anydifference in puzzle solving skills between children tutored by one method versus children tutored by the other. Independent samples of 10 childrenwho were tutored using Method A and 11 children who were tutored using Method B were chosen at random. The Method A children took a mean of40 minutes to solve a certain puzzle with a standard deviation of 6 minutes. The Method B children took a mean of 44 minutes to solve the samepuzzle with a standard deviation of 7 minutes. Assume that the two populations of completion times are normally distributed and that the populationvariances are equal. Construct a 90% confidence interval for the difference u, - u, between the mean puzzle-solving times for Method A children (u) and for Method B children (µ,). Then find the lower limit and upper limit of the 90% confidence interval.Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult alist of formulas.)Lower limit: ||Upper limit:CheckSave For LaterSubmit AssignmerO 2022 McGraw Hill LLC. AlI Rights Reserved. Terms of Use I Privacy Center| AccessibilitAPRtvA6DII1980F7F8F6F5F4escF3F2*-2$8.

There are two independent samples which are method A and method B. We have to construct 90%…

Answered: An educational psychologist is studying…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is greater than the mean of WaxCo. In a random sample of 40 floors of WaxWin and 43 of WaxCo. WaxWin had a mean lifetime of 26.3 and WaxCo had a mean lifetime of 28. The population standard deviation for WaxWin is assumed to be 7.1 and the population standard deviation for WaxCo is assumed to be 6.1. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2. The correct hypotheses are: O Ho: µ1 H2(claim) Ο H: μι μp HA: H1 < µ2(claim) O Ho: µ1 = 2 HA: μι μ (claim) Since the level of significance is 0.05 the critical value is 1.645 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the mean of WaxWin is greater than the mean of WaxCo.…
Research has shown that, for baseball players, good hip range of motion results in improved performance and decreased body stress. A research article reported on a study of independent samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean hip range of motion was 75.5 degrees and the sample standard deviation was 5.8 degrees, whereas the sample mean and sample standard deviation for position players were 79.1 degrees and 7.1 degrees, respectively. Assuming that the two samples are representative of professional baseball pitchers and position players, test hypotheses appropriate for determining if there is convincing evidence that the mean range of motion for pitchers is less than the mean for position players. (Use ? = 0.05. Use ?1 for pitchers and ?2 for position players.) State the appropriate null and alternative hypotheses. Find the test statistic and P-value. (Use SALT. Round your test statistic to one decimal place and…
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B 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_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…
A professor is concerned that the two sections of college algebra that he teaches are not performing at the same level. To test his claim, he looks at the mean exam score for a random sample of students from each of his classes. In Class 1, the mean exam score for 13 students is 75.1 with a standard deviation of 5.8. In Class 2, the mean exam score for 16 students is 82.6 with a standard deviation of 7.3. Test the professor's claim at the 0.02 level of significance. Assume that both populations are approximately normal and that the population variances are equal. Let Class 1 be Population 1 and let Class 2 be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$⁢151 per month with a standard deviation of $16$⁢16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$⁢158 per month with a standard deviation of $19$⁢19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
The average speed of 64 random cars traveling on a certain highway was 80 km/h and the sample standard deviation was 20 km/h. What is the margin of error with a 98% level of confidence for the population average speed? Select one: а. 1.295 b. 2.387 C. 6.64 d. 3.2375 е. 2.656 f. 5.9675
Two universities in the same state are bitter rivals. Each university believes that it's students are more physically fit than the students at the other university. To test the claim that there is a difference in the average fitness levels of students at the two universities, 36 randomly selected students at the first university were surveyed and exercised for a mean of 2.9 hours per week. A random sample of 38 students at the second university was also surveyed, and exercised for a mean of 2.7 hours per week. Assume that the population standard deviation for hours of exercise at the first university is known to be 1.1 hours per week and the population standard deviation for the second university is known to be 1.0 hour per week. Use a 0.05 level of significance to perform a hypothesis test to determine if there is a difference in the average fitness levels of students at the two universities. State the null and alternative hypothesis Determine which distribution to use for the test…
(a) Using the information above, part of the ANOVA table is: Source of variation Sum of squares Region I error II Find the values of (round off to two decimal points) (1) I (11) II
A professor is concerned that the two sections of college algebra that he teaches are not performing at the same level. To test his claim, he looks at the mean exam score for a random sample of students from each of his classes. In Class 1, the mean exam score for 12 students is 80.8 with a standard deviation of 5.7. In Class 2, the mean exam score for 14 students is 84.5 with a standard deviation of 4.9. Test the professor's claim at the 0.01 level of significance. Assume that both populations are approximately normal and that the population variances are equal. Let Class 1 be Population 1 and let Class 2 be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.

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