Question

The General Social Survey is an annual survey given to a random selection of about 1500 adults in the United States. Among the many questions asked are "What is the highest level of education you've completed?" and "If you're employed full-time, how many hours do you spend working at your job during a typical week?"

In a recent year, 1099 respondents answered both questions. The summary statistics are given in the chart below. (The sample data consist of the times, in hours per week, that were given by the respondents.)

To decide if there are any differences in the mean hours per week worked by these different groups, we can perform a one-way, independent-samples ANOVA test. Such a test uses the statistic

F= Variation between Samples/ Variation within the samples

≈F 2.66For the data from the survey,

Give the numerator degrees of freedom of this /F/ statistic.
Give the denominator degrees of freedom of this /F/ statistic.
From the survey data, can we conclude that at least one of the groups differs significantly from the others in mean hours worked in a typical week? Use the 0.05 level of significance.	Yes	No

.

Groups	Sample size	Sample mean	Sample variance
Less than h.s.	253	42.0	93.5
High school	267	41.0	92.7
Bachelor's	288	42.5	104.2
Graduate	291	43.3	95.6

Answer 1

Given,

Groups	Sample size	Sample mean	Sample variance
Less than h.s.	253	42.0	93.5
High school	267	41.0	92.7
Bachelor's	288	42.5	104.2
Graduate	291	43.3	95.6

F ≈ 2.66

That means we have

t=4

N=1099

Answer(1): The numerator degrees of freedom of this /F/ statistic is equal to the degree of freedom for between groups variation i.e. t-1

= 4-1

                                                          = 3

Answer(2): The denominator degrees of freedom of this /F/ statistic is equal to the degree of freedom for within groups variation (Error) i.e. N-t

= 1099-4

                                                          = 1095

Answer(3):

We have to test

H0: there is no difference in groups mean hours worked in a typical week.

H1: At least one of the groups differs significantly from the others in mean hours worked in a typical week.

Now we have degree of freedom for F statistic, so we can find the table value of F for (5,1095) at 0.05 level of significance.

Table value of F at 0.05 level of significance, F(3,1095) = 2.61

Here,

Fcalculated > Ftable , which suggest that we have enough evidence against H0 to reject it at 0.05 level of significance.

Also, the p-value for the above test is 0.047 which is less than 0.05 and confirms the above result.

Hence, we conclude that at least one of the groups differs significantly from the others in mean hours worked in a typical week.

The answer is Yes.