Question

​ A survey of 200 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. Suppose that based on the survey, 70 of the 200 students responded "yes".

a) What is the value of the sample proportion, p ?

b) What is the standard error of the sample proportion?

c) Construct an approximate 95% confidence interval for the true proportion π (pi) by taking +/- 2 SEs (standard errors) from the sample proportion.

d) What would be the minimum sample size for this data with a desire sampling error of 2% (that is, e = +/- 0.02) at a 95% confidence?

Answer 1

The sample proportion is 0.35, the standard error is 0.036, the confidence interval is approximately (0.278, 0.422), and the minimum sample size for a desired sampling error of 2% is calculated using the formula (1.96^2 * 0.35 * (1-0.35)) / (0.02^2).

Step-by-step explanation:

a) The value of the sample proportion, p, is the number of students who responded 'yes' divided by the total number of students surveyed. In this case, p = 70/200 = 0.35.

b) The standard error of the sample proportion is calculated using the formula: SE = sqrt[(p*(1-p))/n], where p is the sample proportion and n is the sample size. In this case, SE = sqrt[(0.35*(1-0.35))/200] ≈ 0.036.

c) To construct an approximate 95% confidence interval for the true proportion, π, we can take +/- 2 SEs from the sample proportion. The confidence interval is given by: (p - 2SE, p + 2SE). In this case, the confidence interval is approximately (0.35 - 2(0.036), 0.35 + 2(0.036)), which is approximately (0.278, 0.422).

d) To calculate the minimum sample size needed with a desired sampling error of 2% and a 95% confidence, we can use the formula: n = (Z^2 * p * (1-p)) / E^2, where Z is the z-score corresponding to the desired confidence level, p is the estimated proportion from the pilot survey, and E is the desired sampling error. In this case, the minimum sample size would be approximately (1.96^2 * 0.35 * (1-0.35)) / (0.02^2).