3. To estimate the effectiveness of a tutoring center its director chose 20 students at random to comparetheir test scores before using the center with their test scores after using the center. The correctformula to se to construct a confidence interval for the mean difference in scores is(a) 표 ± to/2(fa)(b) d± ta/2(c) (포1 - E2) ± ta/2/s(금 + ), s(n, –1)s}+(n2-1)«?n1+n2-2(d) (1 - ī2) ± Za/2V n(c) (fi – P2) ± Za/2/1212

We want to tell you which one is the correct choice.When population variance are unknown then we use tcritical We want to find the confidence interval for difference of mean Hence option (c) is the correct choice.

Answered: 3. To estimate the effectiveness of a…

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 sample of n=300 male college freshmen has a mean weight of 160 with a standard devation of 11. how many of the 300 observed weights fall within (A)2 standard deviations of the mean? (B) 3 standard deviations of the mean? (C) 5/2 deviations of the mean? Show the interval in each case.
A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat type II diabetes. The efficacy of the medication will be determined by the blood glucose readings obtained from the patients. If the researcher wants a margin of error less than or equal to 5 mg/dL and the standard deviation for blood glucose readings among type II diabetics was previous documented as 15 mg/dL, how many patients should be recruited for each group of individuals in the study assuming a 95% confidence interval will be used to quantify the mean difference in blood glucose scores between the control group and the treatment group? [1] n for the treatment group = 32 and n for the control group = 13 [2] n for the treatment group = 33 and n for the control group = 33 [3] n for the treatment group = 70 and n for the control group = 70 [4] n for the treatment group = 1729 and n for the control group = 1729
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A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.07 years, with sample standard deviation s = 0.88 years. However, it is thought that the overall population mean age of coyotes is ? = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use ? = 0.01. (a) What is the level of significance?State the null and alternate hypotheses. H0: ? = 1.75 yr; H1: ? > 1.75 yrH0: ? > 1.75 yr; H1: ? = 1.75 yr H0: ? = 1.75 yr; H1: ? ≠ 1.75 yrH0: ? < 1.75 yr; H1: ? = 1.75 yrH0: ? = 1.75 yr; H1: ? < 1.75 yr (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The Student's t, since the sample size is large and ? is known.The standard normal, since the sample size is large and ? is unknown. The standard normal, since the sample size is large and ? is known.The Student's t, since the sample size is…
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