Question

In: Statistics and Probability

a. Develop a null and alternative hypothesis for a test of whether or not the mean...

a. Develop a null and alternative hypothesis for a test of whether or not the mean of the data representing wives’ satisfaction scores

and husbands’ satisfaction scores are statistically different. Find descriptive statistics on the three relevant variables in the table.

Highlight the means, standard deviations, and observations.

TABLE C9-2: Wife and Husband Feedback Scores on 1-7 Scale + Same Day or Not
Appraisal Survey Wife's Score Husband's Score Same Day
1 4 5 Y
2 3 2 Y
3 1 5 N
4 7 3 N
5 4 3 Y
6 3 6 Y
7 5 3 Y
8 4 7 N
9 5 5 Y
10 1 4 N
11 7 7 Y
12 2 6 N
13 2 5 N
14 5 2 Y
15 3 4 Y
16 5 7 Y
17 3 4 Y
18 1 4 N
19 4 6 Y
20 7 7 Y

Solutions

Expert Solution

Sol:

to get descriptive statsictics in excel

Install analysis tool pak and then go to

Data>data anlayhsis>descriptive stats

You will get

Wife's Score Husband's Score
Mean 3.8 Mean 4.75
Standard Error 0.426738 Standard Error 0.369031
Median 4 Median 5
Mode 4 Mode 5
Standard Deviation 1.90843 Standard Deviation 1.650359
Sample Variance 3.642105 Sample Variance 2.723684
Kurtosis -0.65341 Kurtosis -1.06931
Skewness 0.216058 Skewness -0.10245
Range 6 Range 5
Minimum 1 Minimum 2
Maximum 7 Maximum 7
Sum 76 Sum 95
Count 20 Count 20

. Develop a null and alternative hypothesis for a test of whether or not the mean of the data representing wives’ satisfaction scores

and husbands’ satisfaction scores are statistically different.

Ho:mu1=mu2

Ha:mu1 not = mu2

alpha=0.05

perform t test for difference in means in excel

In excel

go to data >data analysis>t test assuming unequal variance

You will get

t-Test: Two-Sample Assuming Unequal Variances
Wife's Score Husband's Score
Mean 3.8 4.75
Variance 3.642105263 2.723684211
Observations 20 20
Hypothesized Mean Difference 0
df 37
t Stat -1.683885151
P(T<=t) one-tail 0.050311787
t Critical one-tail 1.68709362
P(T<=t) two-tail 0.100623575
t Critical two-tail 2.026192463

From table

t=-1.6839

p=0.1006

p>0.05

Fail to reject null hypothesis

Accept null hypothesis.

conclusion:

the mean of the data representing wives’ satisfaction scores

and husbands’ satisfaction scores are NOT  statistically different at 5% level of significance.


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