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Question 28 3 pts Suppose that the graduation rate in a school district is 85 percent. A random sample of students from the d

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Answer #1

28)

Null Hypothesis : Mean = 85%

Alternate Hypothesis : Mean != 85%

Standard deviation of 85, 1s and 15, 0s is 0.3589

Standard error of sample size 100 = standard deviation/ sqrt(100) = 0.03589

Sample mean = 90%

Sample statistic (z) = (90% - 85%)/standard error = 1.4

So, Option D looks good.

29)

Mean = 175

Standard deviation = 50

z score = -0.5

Actual score = Mean + z score*standard deviation = 175 - 0.5*50 = 150

Option D looks good.

30)

sample size = 25

sample mean = 30

sample variance = 100

degree of freedom (df) = n - 1 = 25 - 1 =24

95% confidence interval

t stat for 95% confidence interval and df as 24 = 2.0639

sample standard deviation = sqrt(sample variance) = sqrt(100) = 10

sample standard error = sample standard deviation/sqrt(n) = 10/sqrt(25) = 2

confidence interval

(mean - t*standard error, mean + t*standard error)

(30 - 2.0639*2, 30 + 2.0639*2)

(25.87 , 34.13)

Option D looks good.

Please like, if you understood. Thanks.

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