Question

1) An economist reports that 700 out of a sample of 2,800 middle-income American households actively participate in the stock market.[You may find it useful to reference the z table.]
  a. Construct the 90% confidence interval for the proportion of middle-income Americans who actively participate in the stock market. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)  

b. Can we conclude that the percentage of middle-income Americans who actively participate in the stock market is not 28%?

Yes, since the confidence interval contains the value 0.28.

Yes, since the confidence interval does not contain the value 0.28.

No, since the confidence interval contains the value 0.28.

No, since the confidence interval does not contain the value 0.28.

Answer 1

1)

Solution :

Given that,

n = 700

x = 2800

Point estimate = sample proportion = = x / n = 700 / 2800 = 0.25

1 - = 1 - 0.25 = 0.75

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.250 * 0.750) / 2800)

= 0.013

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.250 - 0.013 < p < 0.250 + 0.013

0.237 < p < 0.263

The 90% confidence interval for the population proportion p is : (0.237 , 0.263)

Yes, since the confidence interval does not contain the value 0.28.