Question

A large employer would like to estimate the proportion of employees who are dissatisfied with their jobs. They take a random sample of 60 employees and survey them on their current feelings toward their job. Twenty-four of the employees surveyed stated they were moderately to very dissatisfied with their current position. Calculate a 99% confidence interval that estimates the true proportion p, of all employees who are dissatisfied. Group of answer choices 99%CI(0.261, 0.539) 99%CI(0.296, 0.504) 99%CI(0.237, 0.563) 99%CI(0.276, 0.524)

Answer 1

Answer:

99%CI (0.237, 0.563)

Step-by-step explanation:

The z-score for a 99% confidence interval is Z = 2.576

The proportion of workers in the sample who are dissatisfied with their position is given by the number of people who answered that "they were dissatisfied", divided by the total sample of n = 60 employees.

 $p=\frac{24}{60}=0.4$

The confidence interval for a proportion 'p' is:

$p\pm Z*\sqrt{\frac{p*(1-p)}{n}}\\p\pm 2.576*\sqrt{\frac{0.4*(1-0.4)}{60}} \\L=0.4-0.163=0.237\\U=0.4+0.163=0.563$

The 99% confidence interval is CI (0.237, 0.563).

Answer 2

Answer:

99% confidence interval that estimates the true proportion p, of all employees who are dissatisfied is [0.237 , 0.563].

Step-by-step explanation:

We are given that a large employer would like to estimate the proportion of employees who are dissatisfied with their jobs. They take a random sample of 60 employees and survey them on their current feelings toward their job. Twenty-four of the employees surveyed stated they were moderately to very dissatisfied with their current position.

Firstly, the pivotal quantity for 99% confidence interval for the true proportion p is given by;

     P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of employees who stated that they were moderately to very dissatisfied with their current position in a sample of 60 = $\frac{24}{60}$ = 0.40

           n = sample of employees = 60

           p = true proportion

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the true proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                                    significance level are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p =[$\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

 = [ $0.40-2.5758 \times {\sqrt{\frac{0.40(1-0.40)}{60} } }$ , $0.40+2.5758 \times {\sqrt{\frac{0.40(1-0.40)}{60} } }$ ]

 = [0.237 , 0.563]

Therefore, 99% confidence interval that estimates the true proportion p, of all employees who are dissatisfied is [0.237 , 0.563].