5.8 3.3 2.8 95.6 5.0 3.4Suppose a study reported that the average person watched 5.51 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 1%significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the study? (Note:x= 4.287 hours and s= 1.244 hours.)3.2 5.7 2.94.3 6.2 5.73.5 3.9 3.0Click here to view a partial table of values of tSet up the hypotheses for the one-mean t-test.Ho PHa PThe test statistic is t=(Round to two decimal places as needed.)The P-value is(Round to four decimal places as needed.)the value reported in the study.sufficient evidence to conclude that the amount of television watched by the average person isthe null hypothesis. The data

Answered: Image /qna-images/answer/51ccc2e1-4a88-4ceb-b148-4e11da3928e2.jpg

Answered: Suppose a study reported that the…

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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Which of the following z-score values represents the location farthest from the mean? Question 21 options: z = +0.50 z = +1.00 Z=-1.00 z = -2.00
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 98.35, y 99.1, r= 0.879, P-value = 0.000, and y = 10.73 + 0.9x, where x represents the IQ score of the wife. Find the best predicted value of y given that the wife has an IQ of 106? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is. (Round to two decimal places as needed.)
The control department of a light bulb manufacturer randomly picks 4400 light bulbs from the production lot every week. The records show that, when there is no malfunction, the defect rate in the manufacturing process ( due to imperfections in the material used) is 1%. When 1.25% or more of the light bulbs in the sample of 4400 are defective, the control unit calls repair technicians for service. Answer the following.
Before the furniture store began its ad campaign, it averaged 182 customers per day. The manager is investigating if the average has changed since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 194, 170, 172, 180, 151, 169, 179, 181, 167, 170, 194 Assuming that the distribution is normal, what can be concluded at the a = 0.01 level of significance? %3D a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Но: ? v Select an answer v H1: ? v Select an answer ♥ c. The test statistic ? v (please show your answer to 3 decimal places.) %3D d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? va f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... the null hypothesis. O The data suggest that the population mean number of customers since the ad campaign began is not significantly different from 182 at a = 0.01, so there is…
In a study conducted by the U.S. Department of Health and Human Services, a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Find the P-value and state a conclusion. Round the answer to four decimal places.
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 98.91, y = 99, r= 0.937, P-value = 0.000, and y = 5.67 + 0.94x, where x represents the IQ score of the wife. Find the best predicted value of y given that the wife has an IQ of 95? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)
A study was performed to determine the percentage of people who wear life vests while out on the water. A researcher believed that the percentage was different for those who rode jet skis compared to those who were in boats. Out of 500500 randomly selected people who rode a jet ski, 85%85% wore life vests. Out of 250250 randomly selected boaters, 90.4%90.4% wore life vests. Using a 0.050.05 level of significance, test the claim that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat. Let jet skiers be Population 1 and let boaters be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0Ha: p1=p2: p1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯p2 Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places. 3) draw a conclusion and form a decision (word it like "we fail to reject the null hypothesis and…
Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300 said they were current smokers. b. Estimate the variability in such sample proportion by finding SE (p).
The mean BMI in patients free of diabetes was reported as 28.2. An investigator conduction the study hypothesizes that the BMI in patients free of diabetes is higher. Based on the data, is there evidence that the BMI is sigificantly higher than 28.2? Use a 5% level of significance. 25 . 27 . 31 . 33 . 26 . 28 . 41 . 24 . 32 . 35 . 40 .
A random sample of 50 plants was selected from a field containing one-year-old plants of a variety of guayule, a plant species yielding rubber. Of the 50, 25 were classified as Normal (N), 14 were classified as Off-type (O) and 11 as Aberrant (A). The data on the following page show the percentage rubber content from each plant. Normal Off-type Aberrant 6.34 6.40 5.70 4.84 5.01 6.20 4.25 8.90 6.93 6.38 6.68 7.10 6.42 7.26 7.29 6.30 7.01 6-84 7.12 6.83 7.31 7.25 6.86 6.68 5.82 4.46 7.70 5.90 5.24 5.58 6.36 7.10 6.81 7.32 6.48 5.88 5.24 5.51 7.00 6.42 6.20 6.04 5.58 4.72 7.42 7-27 6.43 6-10 6-01 7.90 (a) Test the null hypothesis that there is no difference in mean rubber content between the three types of plant. (b) If the means of the distributions of each type of plant are denoted by N, PO and PA respectively, test the following particular null hypotheses: (i) EN = RA; (ii) Po=(PN + PA).
Suppose a study reported that the average persin watched 3.37 hours of television per day. a random sample of 15 people gave the number of hours of television watched per day shown below. at the 1% significance level, do the data provide sufficent evidence to conclude that the amount of television watched per day last year by the average person is greater than the value reported in the study? what is the test statistic value for this problem? a.0.28 b.-38 c.none of the above d. 0.58 e.-4.295 f..04 g. 1.94 . the proper conclusion for this problem is? a. reject the null hypothesis we have sufficent evidence to prove the average number of hours people watch tv is more than 3.37 hours. b. Do not reject the null hypothesis we have sufficient evidence to prove the average number of hours of TV watched is greater than 3.37 hours c. Reject the null hypothesis and claim the average number of hours people watch TV is less than 3.37 because the P value is near zero d. Do not reject the null…
A dietician wishes to see if a person’s cholesterol level will change if the diet is supplemented by a certain mineral. Six randomly selected subjects were pretested, and then they took the mineral supplement of a 6 week period. The results are shown in the table below. Can it be concluded that the cholesterol level has changed at a significance level of 0.10? Use the P-value method with your calculator. Assume the variable is approximately normally distributed. You can use your calculator to answer this question but make sure that you clearly write the statements for the null and alternative hypothesis. Be sure that you thoroughly explain your final conclusion using the P-value method. Subject 1 2 3 4 5 6 Before 210 235 208 190 172 244 After 190 170 210 188 173 228

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