An experiment is conducted to determine whether intensive tutoring (covering a great deal of material in a fixed amount of time) is more effective than paced tutoring (covering less material in the same amount of time).Two randomly chosen groups are tutored separately and then administered proficiency tests. The data collected and summarized below: Group1 (the intensive group) consisted of 12 students and resulted in an averagegrade of 46.31 with a standard deviation of 6.44. group 2 (the paced group) consisted of 10 students and resulted in an average grade of 42.79 with a standard deviation of 7.52. Assume that the population variances ofboth groups are different. To test the claim that intensive tutoring is more effective than paced tutoring, find the degree of freedom and the critical value , and make a decision.df= 17t critical value to 05 = 1.739Fail to reject HO.df= 17t critical value to 05 = 1.782Fail to reject HO.df= 17t critical value to.025 = 1.986Fail to reject HO.df= 17t critical value to.025 = 1.986Reject H0.df= 17t critical value to.05 = 1.739Reject HO.

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Answered: An experiment is conducted to determine…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The National Assessment of Educational Progress (NAEP) is a nationwide assessment of students' proficiency in nine subjects: mathematics, reading, writing, science, the arts, civics, economics, geography, and U.S. history. The main NAEP assessments are conducted annually on samples of students from grades 4, 8, and 12. Suppose in 2000, the science scores for female students had a mean of 146 with a standard deviation of 35. Assume that these scores are normally distributed with the given mean and standard deviation. The normal curve representing this distribution of the scores is symmetrical about a vertical line through . The value 41 is below the mean, and the value 251 is above the mean. The area under the curve between 41 and 251 is approximately . Similarly, 95.44% of the female students scored between and .
Two brands of batteries are tested, and their voltages are compared. The summary statistics follow. For sample 1, the sample size is 27; the population standard deviation is 0.3 volts and the mean is 9.2 volts. For sample 2, the sample size is 30; the population standard deviation is 0.1 volts and the mean is 8.8 volts. Is the mean of sample 1 smaller than the mean of sample 2? Using a significance level of 0.05. a) Identify the claim and state the hypotheses. b) What is the test statistics? Find the critical values (s). c) Make the decision to reject or not reject the null hypothesis and explain why. d) Graph your decision and include all the values. That is, the mean, the standard deviation, the critical value(s), the rejection region. e) Set up the formula with the correct numbers for the 95% confidence interval of the mean number of jobs.
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
An experiment was run to compare three groups. The sample sizes were 28, 33, and 102, and the corresponding estimated standard deviations were 2.7, 2.6 and 4.8. 1) What is the value of the pooled standard deviation? 2) Explain why the answer for part 1 is much closer to the standard deviation of the third group than to either of the other two standard deviations.
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.
Dr. Graham is interested in determining if middle-aged adults use text messaging more or less frequently than the general population. Dr. Graham collects information on text messaging from a random sample of 50 adults ages 25 to 44. Dr. Graham finds that these individuals send or receive an average of 68 text messages per day. Using the population mean (and standard deviation) of 41.5 texts per day (34 texts per day), determine whether adults in this age group use text messaging more than the general public.
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 successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?
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.
A study was conducted for an aviation safety agency in which they surveyed thousands of passengers at airports across Europe. In this study, the "mass" for a passenger includes the weight of the passenger and all of the passenger's carry-on items, including infants without their own seats. For the 5,904 adult male passengers measured in summer, the sample mean and standard deviation for this mass were 88.8 kg and 15.9 kg, respectively. (a) Construct a 95% confidence interval (in kilograms) for the population mean mass for adult male passengers in summer. (Round your answers to three decimal places.) ___to__ kg (b) Write a sentence or two interpreting the confidence interval you found in part (a). With 95% confidence, we can estimate that the population mean mass of male passengers traveling in the summer is contained within the interval. We can estimate that 95% of the male passengers traveling in the summer in the sample have a mass that is contained within the interval.…
Do political science classes require less writing than history classes? The 47 randomly selected political science classes assigned an average of 19 pages of essay writing for the course. The standard deviation for these 47 classes was 5.6 pages. The 56 randomly selected history classes assigned an average of 21.9 pages of essay writing for the course. The standard deviation for these 56 classes was 4.1 pages. What can be concluded at the a = 0.10 level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer v Select an answer v H: Select an answer v Select an answer v Select an answer v b. The test statistic ? v = (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? va e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the final conclusion is that ... O The results are statistically…
Do political science classes require more writing than history classes? The 40 randomly selected political science classes assigned an average of 13.1 pages of essay writing for the course. The standard deviation for these 40 classes was 3.9 pages. The 53 randomly selected history classes assigned an average of 10.8 pages of essay writing for the course. The standard deviation for these 53 classes was 5.6 pages. What can be concluded at the a = 0.05 level of significance? For this study, we should use Select an answer ↑ a. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer H₁: Select an answer Select an answer Select an answer b. The test statistic ? (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? a e. Based on this, we should Select an answer the null hypothesis. f. Thus, the final conclusion is that …..

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