Answer:
d. (0.5457 , 0.8361)
Step-by-step explanation:
From the missing findings recorded in the Excel output; we have:
the sample size to be = 55
count of response = 38
So, the proportion of parents that have the feeling they do spend little time with their children are :
p = [tex]\dfrac{38}{55}[/tex]
p= 0.69
Thus, p = sample mean [tex]\overline x[/tex] = 0.69
The standard error of the proportion p is expressed as:
[tex]S.E = \sqrt{\dfrac{p (1-p)}{n}}[/tex]
[tex]S.E = \sqrt{\dfrac{0.69 (1-0.69)}{55}}[/tex]
[tex]S.E = \sqrt{\dfrac{0.69 (0.31)}{55}}[/tex]
[tex]S.E = \sqrt{\dfrac{0.2139}{55}}[/tex]
[tex]S.E = \sqrt{0.003889}[/tex]
S.E = 0.06236
At 98% confidence interval level, the level of significance = 1 - 0.98 = 0.02
[tex]z_{0.02/2} =2.326[/tex]
Confidence interval = [tex]p \pm z_{0.02/2} \times S.E[/tex]
Confidence interval = [tex]0.69 \pm 2.326 \times 0.062[/tex]
Confidence interval = [tex]0.69 \pm 0.144212[/tex]
Confidence interval = [tex](0.69 - 0.144212 \ , \ 0.69 +0.144212 )[/tex]
Confidence interval = [tex](0.545788 \ , \ 0.834212 )[/tex]
Thus, option d. (0.5457 , 0.8361) is the right answer
