Response of Interest YesSample Size 55Count for Response 38Sample Proportion Confidence Interval Confidence Coefficient 0.95Lower Limit Upper Limit Hypothesis Test Hypothesized Value 0.60Test Statistic P-value (Lower Tail) 0.9156P-value (Upper Tail) P-value (Two Tail) 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% at α=.1? 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. (0.5688, 0.8130)c. None of the answers is correctd. (0.5457, 0.8361)e. (0.5884, 0.7934)

The proportion of working parents who feel they spend too little time with their children due to work commitments is not different from 60%.

Correct option: (C) None of the answers is correct.

Step-by-step explanation:

A two-tailed hypothesis test can be performed to determine whether the proportion of working parents who feel they spend too little time with their children due to work commitments is different from 60%.

The hypothesis is defined as follows:

H₀: The proportion of working parents who feel they spend too little time with their children due to work commitments is 60%, i.e. p = 0.60.

Hₐ: The proportion of working parents who feel they spend too little time with their children due to work commitments is different from 60%, i.e. p ≠ 0.60.

The information provided is:

n = 55

X = 38

Significance level, α = 0.10

p-value for left tail is, P (Z < z) = 0.9156.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected. And if it is more that the significance level then the null hypothesis will not be rejected.

Compute the p-value of the test from the information provided:

p-value = 2 × P (Z > z)

            = 2 × [1 - P (Z < z)]

            = 2 × [1 - 0.9156]

            = 0.1688

So,

p-value = 0.1688 > α = 0.10

Thus, the null hypothesis was failed to be rejected at 10% level of significance.

Hence, concluding that the proportion of working parents who feel they spend too little time with their children due to work commitments is not different from 60%.

The (1 - α)% confidence interval for population proportion is:

CI = p ± z-critical × √[(p (1 - p))/n]

Compute the sample proportion as follows:

p = X/n

  = 38/55

  = 0.691

The critical value of z for 98% confidence level is:

z-critical = 2.33

*Use a z-table for the value.

Compute the 98% confidence interval for population proportion as follows:

CI = p ± z-critical × √[(p (1 - p))/n]

   = 0.691 ± 2.33 × √[(0.691 (1 - 0.691))/55]

   = 0.691 ± 0.009

   = (0.682, 0.700)

Thus, the 98% confidence interval for population proportion of working parents who feel they spend too little time with their children due to work commitments is (0.682, 0.700).

Correct option is (C).

imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation: