Question

Find the critical z value for a confidence level of 94%.

Critical Value= round answer to 2 decimal places)

Out of 300 people sampled, 246 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

I am 99% confident that the proportion of people who have kids is between  and

Out of 300 people sampled, 255 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

I am 95% confident that the proportion of people who have kids is between  and

Answer 1

Solution:

1)

Find the critical z value for a confidence level of 94%.

c = 94% = 0.94

= 1- c = 1- 0.94 = 0.06

  /2 = 0.03 and 1- /2 = 0.970

Search the probability 0.970 in the Z table and see corresponding z value

  = 1.88

Critical value = 1.88

2)

Out of 300 people sampled, 246 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

n = 300

x = 246

Let denotes the sample proportion.

     = x/n   = 246/300 = 0.82

Our aim is to construct 99% confidence interval.

c = 0.99

= 1 - c = 1- 0.99 = 0.01

  /2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

Now , the margin of error is given by

E = /2 *  

= 2.576 * [0.82 *(1 - 0.82)/300]

= 0.057

Now the confidence interval is given by

( - E)   ( + E)

(0.82 - 0.057)   (0.82 + 0.057)

0.763 0.877

I am 99% confident that the proportion of people who have kids is between  0.763 and  0.877

3)

Out of 300 people sampled, 255 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

n = 300

x = 255

Let denotes the sample proportion.

     = x/n   = 255 /300 = 0.85

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [0.85*(1 - 0.85)/300]

= 0.040

Now the confidence interval is given by

( - E)   ( + E)

(0.85 - 0.040)   (0.85 + 0.040)

0.810 0.890

I am 95% confident that the proportion of people who have kids is between  0.810 and  0.890