The grade point average of 4 randomly selected college students is recorded at the end of the fall and springsemesters of their senior year, as follows:Student1234Fall2.71.73.02.1Spring2.20.902.82.2Construct a 95% confidence interval for the mean difference in GPALower bound =(use two decimal places)Upper bound(use two decimal places)

We have to create 95% confidence interval for difference in mean for paired data using t statistics.

Answered: The grade point average of 4 randomly…

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 simple random sample of 5 recently hospitalized migraine sufferers were given a new treatment; the results are listed below. Construct the 90% confidence estimate of the mean difference in the number of migraines per month before and after the treatment. Summary of # of migraines before and after treatment (5/3, 9/9, 7/4, 5/2, 3/4) Subject A Subject B Subject C Subject D Subject E # of migraines before 5 9 7 5 3 # of migraines after 3 9 4 2 4 **Let the first population be of the "before" results and the second population be of the "after" results.** The right hand critical value is . Round to 2 decimal places. The error is . Round to 2 decimal places. The mean difference in migraines per month is between and .
An experiment was conducted on rodents to determine if the drug Haldol influenced the amount of "rough play." Five groups were included in the experiment. Subsequent behavior in terms of mean "rough play" behaviors was recorded. The output and box plot below, display the mean differences from baseline (i.e., after the drug was administered compared to before drug administration). What are the null and alternative hypotheses of this test? Summary of Roughplay differences Scd. Dev. Haldol dese Mean Freq. -.63158419 13.290618 19 Vehicle | 0.025 mg i 0.05 mg/ 1 0.1 mg/k -12.754389 13.783537 0.2 mg/k I 18 1.1481501 20.447041 5.0499951 16.035879 20 19 -20.S5 16.431064 20 Total -5.6631968 1B.558994 96 Analysis of Variance Frob F (p-value) Source SS df KS F Test- Statistic 8999.32832 4 2249.33203 9.63 0.0000 Between groups Within groups 91 260.682623 23722.1178 Tozal 32721.4461 95 344.436275 chif (4) 4.1710 Prob>chiz = 0.303 = Bartlett's test for equal variances: Roeghpiay diferences -50 Ho: 0…
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Suppose that we are interested in comparing the academic success of college students who belong to fraternal organizations and the academic success of those who do not belong to fraternal organizations. Random samples of size 40 are taken from each population where the variances and standard deviations are known. Sample Size Mean GPA Variance (2) 2.03 2.21 Fraternity member (f): 40 Nonmembers (n): 40 .46 .35 Construct a 96% confidence interval estimate for the difference between the mean GPAS (uf- un). Interpret this interval estimate. What can you conclude about the mean GPAs of the two groups?
na test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.4 and a standard deviation of 19.8. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean µ? Omg/dL
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Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.16 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.90 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let 4, denote the mean weight lost on the low-carb diet and u, denote the mean weight lost on the low-fat diet. Use the a = 0.01 level and the TI-84 Plus calculator. Compute the -value. Round the answer to four decimal places. Determine whether to reject H0 State a conclusion
Which samples show unequal variances? Use a = .10 in all tests. Show the critical values and degrees of freedom clearly and illustrate the decision rule. s1 = 10.2, n1 = 22, s2 = 6.4, n2 = 16, two-tailed test s1 = .89, n1 = 25, s2 = .67, n2 = 18, right tailed test s1 = 124, n1 = 12, s2 = 260, n2 = 10, left-tailed test
K Do freshmen spend significantly more time on social media than other class ranks? A recent study investigated the amount of time per day freshmen, sophomores, juniors, and seniors spent on social media while doing schoolwork. The students surveyed were U.S. residents from a 4-year, public, primarily residential institution. The accompanying computer output shows the mean time each cohort spent on social media and an ANOVA table. Construct a 95% confidence interval to compare the population mean time spent per day on social media between freshmen and seniors. Click the icon to view the summary statistics and ANOVA table. The 95% confidence interval is. (Round to one decimal place as needed.) View an ex NOV 9 Summary Statistics and ANOVA Table 6 n tv Mean Std. Dev. Source DF Class Error Total F6 FR 443 62.5 71.6 SS 99509 3 1562 7927979 1565 8027488 Print SO 330 56.5 68.5 MS 33170 5076 JU 383 72.0 80.9 Done 8 F 6.53 FB SE 410 50.4 62.9 P 0.000 ZA 9 F9 X k answer
Use Excel’s Data Analysis tool to conduct a multiple regression, using Risk of Stroke as the dependent variable, and all the rest of variables as independent variables. Choose confidence level at 99%. Risk of Stroke (%) Age Pressure Smoker (Yes=1) 12 57 152 0 24 67 163 0 13 58 155 0 56 86 177 1 28 59 196 0 51 76 189 0 18 56 155 1 31 78 120 0 37 80 135 1 15 78 98 0 22 71 152 0 36 70 173 1 15 67 135 1 48 77 209 0 15 60 199 0 36 82 119 1 8 66 166 0 34 80 125 1 3 62 117 0 37 59 207 1 Interpret the multiple coefficient of determination R-square. Compare the R-square of the simple linear regression and the R-square of the multiple regression, what does the differences in R-squares tell us? What does F-test for significance tell us?
True or False A 95% confidence level in a mean number of minutes that it takes for an amublance to arrive after a 911 call is (4.2, 7.4). Then there is 95% chance that a randomly selected emergency will have an ambulance response time between 4.2 and 7.4 minutes.

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