Answer:
Step-by-step explanation:
This is a test of 2 population proportions. Let 1 and 2 be the subscript for the women that responded positively at the earlier and later survey. The population proportion of women's positive responses at the earlier and later survey would be p1 and p2 respectively.
p1 - p2 = difference in the proportion of women's positive responses at the earlier and later survey.
The null hypothesis is
H0 : p1 = p2
p1 - p2 = 0
The alternative hypothesis is
Ha : p1 > p2
p1 - p2 > 0
it is a right tailed test
Sample proportion = x/n
Where
x represents number of success
n represents number of samples
For earlier survey,
x1 = 2010
n1 = 3000
p1 = 2010/3000 = 0.67
For later survey ,
x2 = 1530
n2 = 3000
p2 = 1530/3000 = 0.51
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (2010 + 1530)/(3000 + 3000) = 0.59
1 - pc = 1 - 0.59 = 0.41
z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)
z = (0.67 - 0.51)/√(0.59)(0.41)(1/3000 + 1/3000)
z = 12.6
From the normal distribution table, the probability value corresponding to the z score is less than 0.00001
Therefore,
p < 0.00001
By using the p value,
Since 0.05 > the p value, we would reject the null hypothesis. Therefore, at
the .05 level, can we can conclude that women think men are less kind, gentle, and thoughtful in the later survey compared with the earlier one