6.78 In a recent survey, the total sleep time per night among college students w
ID: 3305174 • Letter: 6
Question
6.78 In a recent survey, the total sleep time per night among college students was approximately normally distributed with mean 6.78 hours and standard deviation = 1.24 hours. You plan to take an SRS of size n = 150 and compute the average total sleep time. Now you want to use a sample size such that about 95% of the averages fall within ±9 minutes (0.1500 hours) of the true mean = 6.78. of the true mean - 6.8,ote the averae total sle time. Nou (a) Based on the fact that for n-150, 2 -0.202, should the sample sze be larger or smaller than 150? Explain. The sample size should be larger than 150, because we are looking for a smaller standard deviation. The sample size should be smaller than 150, because we are looking for a larger standard deviation. The sample size should be larger than 150, because we are looking for a larger standard deviation. The sample size should be smaller than 150, because we are looking for a smaller standard deviation (b) Using the 95 part of the 68-95-99.7 rule, what standard deviation of x do you need such that 95% of all samples will have a mean within 9 minutes of ? (Round your answer to four decimal places.) 101 c) Using the standard deviation you calculated in part (b), determine the number of students you need to sample. (Round your answer up to the nearest student.) 274 x studentsExplanation / Answer
(B)
0.15 = 1.96*sigma
sigma = 0.15/1.96 = 0.0765
Hence required sigma = 0.0765
(C)
n = (1.24/0.0765)^2 = 262.5264
Hence required sample size = 263
Simple calculations are
mean = 6.78 and std. dev. = 1.24
n = 150
ME = z * sigma/sqrt(n)
0.15 = 1.96 * 1.24/sqrt(n)
n = (1.96*1.24/0.15)^2 = 262.5264
Hence in order to achieve required CI, we need 263 sample size.
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