Data was accessed for a large group of students, considered to be the population

Question

Data was accessed for a large group of students, considered to be the population, whose mean sleep time was 7.1 hours. One hundred random samples of size 40 were taken, and each sample was used to set up a 95% confidence interval for the population mean sleep time, and to test the hypothesis that population mean sleep time equals 7.1 Which one of the following is most plausible? a) 4 intervals contained 7.1 and 4 tests rejected the null hypothesis b) 4 intervals contained 7.1 and 96 tests rejected the null hypothesis c) 96 intervals contained 7.1 and 4 tests rejected the null hypothesis d) 96 intervals contained 7.1 and 96 tests rejected the null hypothesis a) Construct a 95% confidence interval for the peculation mean dice roll, if the mean of a sample of 16 dice is 3.75: the interval extends from _____ to ______. (Report the interval's endpoints to one decimal place.) b) Does your interval contain the true population mean. 3.5? i) yes, clearly ii) yes. but just barely iii) no, not quite iv) no, not at all Cost of coronary bypass surgery at a sample of 9 local hospitals had a mean of 24 thousand dollars. Assume the population standard deviation be 6 thousand dollars. a) Set up a 90% confidence interval for population mean coat the interval extends from ____ thousand dollars to ____ thousand dollars. (Report values to two decimal places.) b) Based on your confidence interval, is 30 thousand dollars a plausible value for the population mean cost? i) yes ii) no iii) borderline Suppose someone wants to test a claim that the overall mean cost of coronary by-pass surgery is 30 thousand dollars. Winch one of these is the correct formulation of the null hypothesis in this case? a) H_0: mu = 24 b) H_0: mu notequalto 24 c) H_0: X = 24 d) H_0: x notequalto 24 e) H_0: mu = 30 f) H_0: x = 30 h) H_0: x notequalto 30

Explanation / Answer

Question-1

Option-c is the answer.

Question-2

Standard deviation=sqrt((1/6)*(1-1/6)/16)=0.093

95% CI=(3.7 , 3.8)

One-Sample T

N    Mean   StDev SE Mean       95% CI

16 3.7500 0.0930   0.0233 (3.7004, 3.7996)

Question-3

Part-a

90% CI= (20.71, 27.29)

One-Sample Z

The assumed standard deviation = 6

N   Mean SE Mean      90% CI

9 24.00     2.00 (20.71, 27.29)

Part-b

No, as this is not in the interval.

Question-4

H0: =30

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