Inferences for p - Population Proportion

Inferences for p - Population Proportion

1.

a)

$99\%$ C.I. for p - proportion of daughters who are taller than their mothers.

n = 30, $\hat{p} = \frac{18}{30}$

x = 18, where x = number of daughters who are taller than their mothers.

$\hat{p} = 0.60$ , z^* = 2.576

$0.6 \pm 2.576\sqrt{\frac{(0.6)(0.4)}{30}}$

$0.6 \pm 0.23$

(0.37, 0.83)

$99\%$ C.I. between $37\%$ and $83\%$ of daughters are tallerthan their mothers.

b)

$H_{0}: \ p = 0.50$

$H_{A}: \ p > 0.50$

$z = \frac{0.60 - 0.50}{\sqrt{\frac{(0.5)(0.5)}{30}}} = \frac{0.10}{0.091} = 1.10$

p-value: P(z > 1.10) = 0.1357 = 1 - 0.8643

Do not reject at $\alpha = 0.10$ or $\alpha = 0.05$ .

2.

p = proportion of elementary school children with hearing problems in the past. p₀ = 0.8

$H_{0}: \ p = 0.08$

$H_{A}: \ p > 0.08$

n = 400, x = 35, $\hat{p} 0.0875$

$z = \frac{0.0875 - 0.08}{\sqrt{\frac{(0.08)(0.92)}{400}}} = 0.55$

p-value: P(z > 0.55) = 0.2912

No evidence against the null hypothesis. Evidence from the sample suggests no signifigant increase.

3.

p = proportion of trucks with falty brakes

n = 40, x = 14, $\hat{p} = \frac{14}{40} = 0.35$

a)

$90\%$ C.I. for p: z^* = 1.645

$0.35 \pm 1.645\sqrt{\frac{(0.35)(0.65)}{40}}$

$0.35 \pm 0.124$

(0.226, 0.474)

The $90\%$ confidence interval tells us that between $22.6\%$ and $47.4\%$ of trucks have faulty brakes.

b)

m = 0.05, $90\%$ C.I.: z^* = 1.645, use $\hat{p} = 0.35$

$n = \frac{(1.645)^{2}(135)(0.65)}{(0.05)^{2}} = 246.2$ , need to test at least 247 trucks.

4.

p = proportion of crayfish in the lake which contain more than 9 parts per billion of mercury.

a)

m = 0.04, $95\%$ C.I.: z^* = 1.96

We do not know p, so we will use p = 0.5.

$n = \frac{(1.96)^{2}(0.5)(0.5)}{(0.04)^{2}} = 600.25$ , need to analyze at least 601 crayfish.

b)

Let p = 0.70, m = 0.04, $95\%$ C.I.: z^* + 1.96

$n = \frac{(1.96)^{2}(0.7)(0.3)}{(0.04)^{2}} = 504.21$ , need to analyze at least 505 crayfish.

Jonathan Payne
1999-04-06