A sample of 26 randomly selected college students at a large university reports
ID: 3053999 • Letter: A
Question
A sample of 26 randomly selected college students at a large university reports getting an average of 5.8 hours of sleep last night with a standard deviation of 2.12 hours. Assume that the sample is normally distributed.
A. Verify that it is reasonable to use the t-distribution to perform the necessary calculations of the confidence interval and significance test for the average amount of sleep students at this university got last night.
B. Create a sampling distribution of all samples with size 26 college students based of your sample.
C. Determine the probability that a sample of size 26 college students will be less than 3 hours of sleep on the previous night.
D. Construct a 96% confidence interval for the average amount of sleep students at this university got last night.
E. According to the school paper, the typical college student at this university gets 6.85 hours of sleep. Test at the 10% significance level that the population of university students is significantly different than the reported amount from the school paper based from the sample data.
Explanation / Answer
Answer to the question is as follows:
n = 26, Xbar = 5.8, s= 2.12
A. Yes, it is reasonable to assume this.
Since the n<30 and we have the sample' deviation and not the population deviation it is safe to assume the t-distribution.
B. Sampling distribution will be ~t(5.8, 2.12/sqrt(26) in the form of ~t(Xbar, s/sqrt(n))
C. P(Xbar<3) = P(Z< (3-5.8)/(2.12/sqrt(26)) = P(Z<-6.74) = ~0
D. 96% interval is :
Xbar+/-t*s/sqrt(n) = 3+/- 2.485*2.12/sqrt(26) = 1.967 to 4.033
E. Xbar = 6.85
t = (6.85-5.8)/(2.12/sqrt(26)) = 2.526
Since this is more than the critical value of 1.708 at df of 26-1=25, we conclude that we reject Ho.
and say that the 10% significance level that the population of university students is significantly different than the reported amount from the school paper based from the sample data
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