Question

1. A public opinion poll asked 930 randomly selected adults if they felt safer given that the U.S. has nuclear weapons, and 543 of the respondents answered “yes.”

1. Find the sample proportion of adults that answered “yes” in this poll.
2. Use a 90% confidence level to compute the margin of error for the point estimate in part (a) above.
3. State the 90% confidence interval for the population proportion of adults who feel safer given that the U.S. has nuclear weapons.
4. Correctly interpret the meaning of your answer from part (c) above in “plain English.” (In other words, explain what this confidence interval tells us as if you were explaining it to a “non-statistics person.”)

2. A confidence interval for the population proportion is: 0.0641 < p < 0.1860. Find the sample proportion and the margin of error for this (confidence) interval estimate.

3. A confidence interval for the population mean is: 24.6 < μ < 31.2. Find the sample mean and the margin of error for this interval estimate.

Answer 1

1.a) Sample proportion of adults that answered yes in the public opinion poll asking that they feel safer given that U.S has nuclear weapons is given by -

b) Margin of errror (M.O.E) for 90% confidence level is given by -

where, is the critical value of z at 0.10 level of significance for the two tailed test. It can be obtained from the z table by finding the z corresponding to 0.05 area and is equal to 1.65.

n is the sample size = 930

p is the sample proportion = 0.584

So, the value of the margin of errror is given by -

= 0.027

c) 90% confidence interval for the population proportion of adults who feel safer given that the U.S has the nuclear weapon is given by -

= [0.557, 0.611]

d) Thus, we are 90% confident that the true proportion of adults who feel safer given that the U.S has the nuclear weapons is between 0.557 and 0.611

2) For confidence interval - [0.0641, 0.1860]

Lower Limit = Sample proportion - Margin of errror

0.0641 = p - M.O.E _________(1)

Upper limit = Sample proportion + Margin of errror

0.1860 = p + M.O.E ________(2)

Adding equation (1) and (2), we get,

2p = 0.2501

p = 0.12505

Putting the value of p obtained in equation (1) or (2), we get,

M.O.E = 0.1860 - 0.12505 = 0.06095

3) For confidence interval - [24.6, 31.2]

Lower Limit = sample mean - Margin of errror

24.6 = - M.O.E __________(1)

Upper limit = sample mean + Margin of errror

31.2 = + M.O.E _________(2)

Adding eqn (1) and (2), we get,

Putting the value of in equation (1), we get,

M.O.E = 3.3

We were unable to transcribe this image

We were unable to transcribe this image

M.O.E = 1.65 * 0.584(1 – 0.584) 930

We were unable to transcribe this image

We were unable to transcribe this image

We were unable to transcribe this image

We were unable to transcribe this image

We were unable to transcribe this image

We were unable to transcribe this image

We were unable to transcribe this image