Question

Confidence Interval for p: Suppose that 60 out of 100 sampled FIU voting-eligible students, support the candidate Joanna in the upcoming election.

a) What is your “best-guess” estimate for the percentage of students in the population who will vote for Joanna?

b) Compute a 95% Confidence interval for the proportion (p) of Joanna’s support in the upcoming election. What is the margin of error?

c) How likely would the data be if p=0.55? Is p=0.55 in the 95% confidence interval?

d) How likely would the data be if p=0.5? Is p=0.5 in the 95% confidence interval?

e) Why are some values of p in the confidence interval and others excluded from the interval?

Answer 1

a) Best guess for the proportion of students who will vote for Joanna: 60%

b)

he following information is provided: The sample size is N = 100, the number of favorable cases is X = 60, and the sample proportion is p=0.6, and the significance level is alpha =0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p = 0.6

Ha: p not= 0.6

This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used.

z_c = 1.96

2=_ ē – po V po(1 – po)/n 0.6 - 0.6 0.6(1 – 0.6)/100

it is then concluded that the null hypothesis is not rejected.

the 95% confidence interval for p is: 0.504 & < 0.6960 i.e 504<p<0.696

c)

he following information is provided: The sample size is N = 100, the number of favorable cases is X = 60X=60, and the sample proportion is pˉ​=0.6, and the significance level is alpha =0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p = 0.55

Ha: p≠​ 0.55

Based on the information provided, the significance level is alpha =0.05, and the critical value for a two-tailed test is z_c = 1.96.

The Z-statistic is computed as follows: z = p – po V po(1 – po)/n 0.6 – 0.55 70.55(1 – 0.55)/100 = 1.005

  Since it is observed that |∣z∣=1.005 ≤ zc​=1.96, it is then concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is different than p0​, at the alpha =0.05 significance level.

the 95% confidence interval for p is 0.504 < p < 0.696

which says, 0.55 belongs to the CI

d)

The following information is provided: The sample size is N =100, the number of favorable cases is X =60, and the sample proportion is \pˉ​=0.6, and the significance level is alpha = 0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p = 0.5

Ha:p≠​0.5

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is z_c = 1.96

The Z-statistic is computed as follows: р — ро 0.6 – 0.5 2 = = 2 V po(1 – po)/n 10.5(1 – 0.5)/100

Since it is observed that ∣z∣=2>zc​=1.96, it is then concluded that the null hypothesis is rejected.

it is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is different than p_0, at the alpha 0.05 significance level.

The 95% confidence interval for p is:0.504<p<0.696.

Which says 0.5 does not belongs to the CI