Question

1. Determine the critical value    that corresponds to a 96% level of confidence.

2. Determine the point estimate () of the population proportion and the margin of error (E) for the given confidence interval. Lower bound: 0.223, upper bound: 0.285  (Round to the nearest thousandth)

3. A survey found 195 of 250 randomly selected Internet users have high-speed Internet access at home. Construct a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home.  (Round to the nearest thousandth)

4. You want to estimate the proportion of adults who have a fear of snakes (ophidiophobia). What sample size should be obtained if you want the estimate to be within 3 percentage points with 95% confidence if you do not use any prior estimate? (Round UP to the nearest whole number)

(a) TRUE/FALSE: As the size of the random sample increases, the margin of error decreases.

(b) TRUE/FALSE: As the level of confidence increases, the margin of error decrease.

Answer 1

1)

Ans:

Using Z table shown below the critical value at 96 % confidence level is z=2.05

Since the confidence interval is 0.119, to 0.381 hence the proportion or the

point estimate is = {0.223+0.285}/2= 0.254

margin of Error = upper limit - point estimate

= 0.285 - 0.254

= 0.031

2)

Ans:

given that n = 250 ,
and x = 195

p^ = x/n = 195/250 = 0.78

then q = 1 - p^ = 1 - 0.78 = 0.22

90% confidence for z is 1.645

confidence interval formula

=> p^ +/- z * sqrt(p*q/n)

=> 0.78 +/- 1.645 * sqrt(0.78*0.22/250)

=> 0.78 +/- 0.043

=> (0.737 , 0.823)

3)

Ans: give ME = 0.03 and level of significance 95%

Z_/2 = Z_0.025 = 1.96

Sample size =

n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.96 / 0.03)² * 0.5 * 0.5 = 1067.11 = 1067

n = sample size = 1067

a)

ans: True

b)

Ans: False

As the level of confidence increases, the margin of error increases.