Question

An economist reports that 238 out of a sample of 700 middle-income American households actively participate in the stock market. [You may find it useful to reference the z table.]
  

a. Construct the 95% 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 38%?

• Yes, since the confidence interval contains the value 0.38.

• Yes, since the confidence interval does not contain the value 0.38.

• No, since the confidence interval contains the value 0.38.

• No, since the confidence interval does not contain the value 0.38.

Answer 1

solution? Given that, n- 700 X 238 P = x = 238 - 0.340 700 point estimate =0.340 L-=0.340 = 0.660 2 At as o% confidence level the z is x-1-95%0=1-o.as=0.05 q=0.99 =0.025 22122 20.025 = 1.9600 Margin of error=E= 2x12x Ô (1-6) - E=1.96oox o.3408 0.660 V 700 E-0.035 At a5 % confidence interval forpis D-E <p s te 0-340-0.035 <p<0. 340 +0.035 0.305 cp<0. 375 [ 0.305 and 0.375)

No since the confidence interval does not contain the value 0.38 option CD) is correct