The dean of a university estimates that the mean number of classroom hours per w
ID: 3270342 • Letter: T
Question
The dean of a university estimates that the mean number of classroom hours per week for full-time faculty is 11.0. As a member of the student council, you want to test this claim. A random sample of the number of classroom hours for eight full-time faculty for one week is shown in the table below. At alphaequals=0.10, can you reject the dean's claim? Complete parts (a) through (d) below. Assume the population is normally distributed.
10.6 9.5. 13.4. 8.9. 5.4. 10.2. 13.7. 8.5
Use technology to find the P-Value.
Decide whether to reject or fail the null hypothesis.
Explanation / Answer
State the hypotheses: The null and alternative hypotheses are as follows:
H0:mu=11 (the mean number of classroom hours per week for full time faculty is 11)
H1:mu=/=11 (the mean number of classroom hours per week for full time faculty is different from 11)
Assumptions and conditions: Randomization condition:-the number of classroom hours for eight full-time faculty were randomly surveyed, so that the classroom hours are mutually independent.
Nearly normal population:-it is given that the population is normally distributed.
The conditions are satisfied, use a Student's t model with (n-1)=(8-1)=7 degrees of freedom to do one-sample t test for the mean.
Test statistic:
From information given, compute xbar=10.025 [xabr=sigma x/n=(10.6+9.5+...+8.5)/8], s=2.686 [s=sqrt{1/n-1 sigma (x-xbar)^2}=sqrt [1/8-1 {(10.6-10.025)^2+(9.5-10.025)^2+...+(8.5-10.025)^2}], n=8
t7=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard devaition and n is sample size.
=(10.025-11)/(2.686/sqrt 8)
=-1.03
Using technology the exact p value is:0.339
Rejection rule:reject null hypothesis if p value is less than alpha=0.10. Here, p value is not less than 0.10.
Conclusion:Fail to reject null hypothesis. There is insufficient evidence to warrant the rejection of the claim.
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