Question

USA Today surveyed fifty-five working parents and asked them if they feel they spend too little time with their children due to work commitments. The findings were recorded In Excel. On the basis of this sample, have we reason to believe that the proportion of working parents who feel they spend too little time with their children due to work commitments is different from 60 to 17 Use the excel output above to answer the following question
What is the 98 confidence interval for the population percentage of working parents who feel they spend too little time with their children due to work commitments?
a. (0.6111, 0.7707)
b. (05084, 0.7934)
c. None of the answers is correct
d. (0.54570.8361)
e. (0.5688.0.8130)

Answer 1

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 = $\dfrac{38}{55}$

p= 0.69

Thus, p =   sample mean $\overline x$ = 0.69

The standard error of the proportion p is expressed as:

$S.E = \sqrt{\dfrac{p (1-p)}{n}}$

$S.E = \sqrt{\dfrac{0.69 (1-0.69)}{55}}$

$S.E = \sqrt{\dfrac{0.69 (0.31)}{55}}$

$S.E = \sqrt{\dfrac{0.2139}{55}}$

$S.E = \sqrt{0.003889}$

S.E = 0.06236

At 98% confidence interval level, the level of significance = 1 - 0.98 = 0.02

$z_{0.02/2} =2.326$

Confidence interval = $p \pm z_{0.02/2} \times S.E$

Confidence interval = $0.69 \pm 2.326 \times 0.062$

Confidence interval = $0.69 \pm 0.144212$

Confidence interval = $(0.69 - 0.144212 \ , \ 0.69 +0.144212 )$

Confidence interval = $(0.545788 \ , \ 0.834212 )$

Thus, option d. (0.5457 , 0.8361) is the right answer