In: Statistics and Probability
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 |
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.