Question

Out of a sample of 200 students, 75 indicated that they preferred chocolate ice cream to vanilla ice cream.

a.	Estimate the value of the population proportion of those who preferred chocolate ice cream. (Round the final answer to 3 decimal places.)

   

Estimated population proportion

   

b.	Compute the standard error of the proportion. (Round the final answer to 4 decimal places.)

Standard error of the proportion

c.	Determine a 90% confidence interval for the population proportion of those who preferred chocolate ice cream. (Round the final answers to 3 decimal places.)

   

  Confidence interval for the population proportion is between  and .

d.	Interpret your findings.

  If 200 such intervals were determined, the population (Click to select)meanproportion would be included in about intervals.

Just Need C and D as my answer for the practice lab is wrong. I want to know ehere im making the mistake.

References

Answer 1

a)

= 75 / 200 = 0.375

b)

Standard error = Sqrt( ( 1 - ) / n)

= Sqrt( 0.375 * 0.625 / 200)

= 0.0342

c)

90% confidence interval for p is

- Z * Sqrt( ( 1 - )/ n) < p < + Z * Sqrt( ( 1 - )/ n)

0.375 - 1.645 * 0.0342 < p < 0.375 + 1.645 * 0.0342

0.319 < p < 0.431

90% CI is (0.319 , 0.431)

Confidence interval for population proportion is between 0.319 and 0.431

d)

If 200 such intervals were determined, the population proportion would be included in

0.319 and 0.431.