Question

A sample of 1006 people found 64% view their local police force favorably.

a.  What is the % chance that more than 60% of people feel this way?

b. What is the chance that between 65% and 68% of people feel the same way?

c. What % chance that more than 70% of local people fell the same way?

At a shooting range, a sample of 714 shots found 13% hit the bullseye.

d. What guess would you say the percentage of all shots hit the bullseye?

e. What numerical range represents the 90% confidence interval for the % of bullseye shots.

f. What numerical range represents the 95% confidence interval for the percentage of shots hitting the bullseye at this shooting range.

Answer 1

n = 1006, = 0.64

(a) The required probability = P(p > 0.60)

Now, standard error = = 0.0151

The Corresponding z value = (0.60 - 0.64)/0.015 = 2.643

Thus, P(p > 0.60) = P(Z > 2.643) = 0.9959

There is a 99.59% chance that more than 60% of people feel this way

(b) Similar to part a, the required probability

= P{(0.65 - 0.64)/0.015 < Z < (0.68 - 0.64)/0.015}

= P(0.667 < Z < 2.667)

= 0.2487

There is a 24.87% chance that between 65% and 68% of people feel the same way

(c) The required probability = P{Z > (0.70 - 0.64)/0.0151}

= P(Z > 3.96) = 0.00004

There is a 0.004% chance that more than 70% of local people feel the same way

In this case,

n = 714, p = 0.13

(d) The guess is that 13% of all shots hit the bull's-eye

(e) Corresponding to 90% confidence interval, the critical z score = 1.645

Thus, the margin of error =

= 0.0207

Thus, the required 90% confidence interval is

(0.13 - 0.0207, 0.13 + 0.0207) = (0.1093, 0.1507)

= (10.93%, 15.07%)

(f) Corresponding to 95% confidence interval, the critical z score = 1.96

Margin of error = 0.0247

The required 95% confidence interval is

(10.53%, 15.47%)