5. Consider the following estimated equation that is used to study the effect ofskipping class on colege GPA:COLGPA- 1.39+412 HSGPA+015 ACT - 083 skipped(011)(.026)(33)(.094)n= 141, R2= 234(1) Using the standard normal approximation, find the 95% confidenceinterval for puA(2) Based on the 95% confidence interval you obtained, can you reject thehypothesis H BOPA = 0.4 against the two-sided alternative at the 5% level?(3) Can you reject the hypothesis H: PPA = 0.1 at the 5% level?(4) Can you reject the hypothesis H,: P 0.8 at the 5% level?

Given Least square equationn = 141R2 = 0.234colGPA=1.39+0.412 hsGPA+0.015 ACT-0.083 skipped

Answered: 5. Consider the following estimated…

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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1.4
17.18 4S overall mortality. The 4S study introduced in Exercise 17.17 also tallied deaths due to any cause. The treatment group experienced 182 deaths (n1 = 2221) and the control group experienced 256 deaths (n2 = 2223). Compare the mortality experience in the two groups in the form of a relative risk and 95% confidence interval.
The Florida State University nursing department is running a Chi Squared analysis and have set their significance at 95%. Based on the following statistical statement, χ2(3, N = 248) = 4.37, p < .51, the researchers should fail to reject the null.
A researcher uses an ANOVA to evaluate the mean difference between two treatments and obtains F(1, 22) = 4.00. If the researcher had used an independent-measures t test to evaluate the data, what t value would be obtained? a. t(23) = 2.00 b. t(23) = 16.00 c. t(22) = 16.00 d. t(22) = 2.00
Suppose that the LFPR in 1968 was 0.58 (or 58 percent). On the basis of the regression results given above, what is the mean LFPR in 1972? Establish a 95% confidence interval for the mean prediction.
Assume that both populations are normally distributed. Test whether u, # µ, at the a= 0.05 level of significance for the given sample data. Population 1 10 Population 2 10 12.9 15.1 2.9 2.6 Test whether u, # H, at the a = 0.05 level of significance for the given sample data. Determine the null and alternative hypothesis for this test. O A. Ho H = H2 H, H, > H2 O B. HoH1 = H2 대# : H OC. Ho H1 # H2 대< H: H O D. Ho H, # H2 H, 4, =2 Detemine the P-value for this hypothesis test. P=(Round to three decimal places as needed.) Should the null hypothesis be rejected? O A. Reject Ho, there is not sufficient evidence to conclude that the two populations have different means. O B. Do not reject Hn, there is sufficient evidence to conclude that the two populations have different means. O C. Reject H,, there is sufficient evidence to conclude that the two populations have different means. O D. Do not reject Hn, there is not sufficient evidence to conclude that the two populations have different…
The following data represent the length of time, in days, to recovery for patients randomly treated with one of two medications to clear up severe bladder infections. Find a 99% confidence interval for the difference µ, - H, between in the mean recovery times for the two medications, assuming normal populations with equal variances. Medication 1 n = = 10 x, = 17 = 1.9 Medication 2 n2 = 14 X2 = Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. = 19 S2 = 1.5 .... . The confidence interval is
A 95% two-sided confidence interval for μ which has been calculated using R turns out to be (0, 1). A 90% two-sided confidence interval based on the same data will contain the value 0.9 (True, cannot tell, or False)
The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65−74 years: n = 116, x = 12.5, and s = 6.35. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use ? = 0.05.)State the appropriate null and alternative hypotheses. H0: ? = 15Ha: ? ≠ 15H0: ? = 15Ha: ? > 15 H0: ? = 15Ha: ? ≤ 15H0: ? = 15Ha: ? < 15 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day.Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. Do not reject…
2) (9.2) Suppose a random sample of 1001 Americans age 15 or older were asked "How much time do you spend eating or drinking per day?" Of the 1001 participants, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hours. Determine a 99% confidence interval for the mean amount of time Americans age 15 or older spend eating or drinking each day.
Suppose you take 4 air quality samples in parts per million: 19.6, 1.88, 12.7, 11.9. Give a 90% confidence interval of the mean air quality? (Also, t(0.95,3) = 2.3534 and t(0.90,4)= 1.5332) (a) (3.45, 16.83) (b) (0, 25) (c) (2.93, 20.10) (d) (5.22, 21.54) (e) None of the above

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