Using the Excel output in Figure 1, what would you conclude testing whether the
ID: 3151346 • Letter: U
Question
Using the Excel output in Figure 1, what would you conclude testing whether the average number of hours students watch TV per week is different from 8 hrs at the 0.05 significance level?
Figure 1: Descriptive statistics on the average number of hours students watch TV per week
Select one:
a. The average number of hours students watch TV per week is equal to 8 hours
b. The population of all students, on average, watch 6.805 hours of TV per week
c. Accept the alternative hypothesis
d. The average number of hours students watch TV per week is different from 8 hours
Using the Excel output in Figure 1 above, what would you conclude to the hypothesis test described in Question 2 at the 0.10 significance level?
Select one:
a. Reject the null hypothesis
b. Fail to reject the null hypothesis
c. Need more information to answer this question
d. Accept the alternative hypothesis
Using the Excel output in Figure 1, what is the lower limit to the 95% confidence interval?
Select one:
a. 5.26
b. 6.25
c. 4.89
d. 4.72
Explanation / Answer
1.
Formulating the null and alternative hypotheses,
Ho: u = 8
Ha: u =/ 8
As we can see, this is a two tailed test.
Getting the test statistic, as
X = sample mean = 6.805
uo = hypothesized mean = 8
n = sample size = 50
s = standard deviation = 5.448106
Thus, z = (X - uo) * sqrt(n) / s = -1.550984147
Also, the p value is
p = 0.120905483
As P > 0.05, we FAIL TO REJECT THE NULL HYPOTHESIS.
Hence,
OPTION a. The average number of hours students watch TV per week is equal to 8 hours [ANSWER]
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2.
OPTION B: b. Fail to reject the null hypothesis [ANSWER]
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3.
Lower = Mean - Margin of Error = 6.805 - 1.548 = 5.26 [ANSWER, A]
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