• Example: A physical therapist wished to estimate, with 99 percent confidence,the mean maximal strength of a particular muscle in a certain group ofindividuals. He is willing to assume that strength scores are approximatelynormally distributed with a variance of 144. A sample of 15 subjects whoparticipated in the experiment yielded a mean of 84.3.

From the provided information, Sample size (n) = 15 Population variance (σ2) = 144 Sample mean (x̄)…

Answered: • Example: A physical therapist wished…

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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A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use a = 0.05 and the following data to test the hypothesis and comment on the value of the commercial. Purchase Rating Individual After Before 1 6 5 6 4 3 7 4 4 3 3 8 8 6 State the null and alternative hypotheses. (Use u, = mean rating after - mean rating before.) O Ho: Hs0 H3: H = 0 O Ho: Hd#0 Hai Hq= 0 O Ho: Hg=0 O Ho: Hd> 0 Calculate the value of the test statistic. (Round your answer to three decimal places.) 1.183 Calculate the p-value. (Round your answer to…
A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use a = .05 and the following data to test the hypothesis and comment on the value of the commercial. Purchase Rating Purchase Rating Individual After Before Individual After Before 1 6. 6. 3 2 6. 4 6. 7 8 7 6. 5 4 4 8 6. 6 a. What are the hypotheses? Ho: ld is Select Ha: ld is select b. Compute d (to 3 decimals). Compute sa (to 1 decimal).
Using traditional methods it takes 8.2 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 26 students and observed that they had a mean of 8.0 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique reduces the training time? Assume the population distribution is approximately normal. Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.
An electronic company's records show that the mean number of circuit.boards rejected per batch is 5.7 with a variance of 1.4. Due to a change in the molding process, factory management has hired you to perform a hypothesis test to check if the variance, o, has increased. To do so, you test a random sample of 10 batches produced using the new process; the number of circuit boards rejected per batch has a sample variance of 2.5. Under the assumption that the number of circuit boards rejected per batch using the new process follows a normal distribution, you will perform a chi-square test. Find x, the value of the test statistic for your chi-square test. Round your answer to three or more decimal places. * = 0
Using traditional methods it takes 8.8 hours to receive a basic driving license. A new license training method using Computer-Aided Instruction (CAI) has been proposed. A researcher used the technique on 17 students and observed that they had a mean of 8.4 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique performs differently than the traditional method? Assume the population distribution is approximately normal. Step 4 of 5 : Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
You are testing the claim that having lights on at night increases weight gain (abstract). A sample of 10 mice lived in an environment with bright light on all of the time and 8 mice who lived in an environment with a normal light/dark cycle is given below. Test the claim using a 10% level of significance. Assume the population variances are unequal and that the weight changes are normally distributed. Give answers to 3 decimal places. Data available at StatKey, choose Mice Wgt Gain-2e data set Light (x₁) 1.71 4.67 4.99 5.33 5.43 6.94 7.15 9.17 10.26 11.67 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer ? V Select an answer V Ha: Select an answer ✓ Based on the hypotheses, find the following: Test Statistic = Dark (₂) 2.27 2.53 2.83 4 4.21 4.6 5.95 6.52 p-value = ? V Select an answer V (Hint: difference in means from Ha) The correct decision is to Select an answer…
please quickly i will give you a thumb!
Use the following information to answer the question. A researcher wants to study different messaging platforms by collecting data on 40 individuals. Half of the individuals only use iMessage while the other half only use What's App. For the individuals who only used iMessage, they sent an average of 280 messages with a sample variance of 170 in the past week. For the individuals who only used What's App, they sent an average of 295 messages with a sample variance of 182 in the past week. Determine if there is a significant difference between the average messages sent between the two groups of individuals. Use a significance level of 1%. What is the null hypothesis? Group of answer choices A. The iMessage group on average sent more messages than the What's App group. B. The iMessage group on average sent less messages than the What's App group. C. There is no difference in the average number of messages sent between the two groups. D. There is a difference in the average…
An intercity bus company is attempting to improve the reliability of its service. One manager claims that the variability (i.e., standard deviation) in arrivals from the posted schedule at the central bus depot is approximately 3 minutes. Another manager disagrees, suspecting that the spread in times of arrival is different. Using variance, what is the population parameter under consideration? What hypotheses should the managers test in this situation?
A researcher decides to measure anxiety in group of bullies and a group of bystanders using a 23-item, 3 point anxiety scale. Assume scores on the anxiety scales are normally distributed and the variance among the group of bullies and bystanders are the same. A group of 30 bullies scores an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. A group of 27 bystanders scored an average of 25.8 with a sample standard deviation of 8 on the anxiety scale. You do not have any presupposed assumptions whether bullies or bystanders will be more anxious so you formulate the null and alternative hypothesis based on that.
In a study performed on comfort food eating during times of stress, it was found that mean rated comfort food consumption level (on a scale from 1-20) for a group of students undertaking their exam period was 14. In comparison, a group of students who were not undertaking exams or examination preparation had a mean rated comfort food consumption level of 10. The pooled estimate of the population variance was 64. What is the estimated effect size (d) of the study? 0.80 0.06 0.50 2.00
A basketball coach believes that the variance of the heights of adult male basketball players is different from the variance of heights for the general population of men. The sample variance of heights, measured in inches, for a random sample of 19 basketball players is 14.22. The sample variance for a random sample of 18 other men is 34.49. Assume that both population distributions are approximately normal and test the coach’s claim using a 0.10 level of significance. Does the evidence support the coach’s claim? Let male basketball players be Population 1 and men in general be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: σ21=σ22: Ha: σ21⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯σ22 Step 2 of 3: Compute the value of the test statistic. Round your answer to four decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.

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