Question

A sample of 307 urban adult residents of a particular state revealed 64 who favored increasing the highway speed limit from 55 to 65 mph, whereas a sample of 179 rural residents yielded 76 who favored the increase. Does this data indicate that the sentiment for increasing the speed limit is different for the two groups of residents?

(a) Test H₀: p₁ − p₂ = 0 versus H_a: p₁ − p₂ ≠ 0 using α = 0.05, where p₁ refers to the urban population. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z =
P-value =

(b) If the true proportions favoring the increase are actually p₁ = 0.24 (urban) and p₂ = 0.42 (rural), what is the probability that H₀ will be rejected using a level 0.05 test with m = 307, n = 179? (Round your answer to four decimal places)

Answer 1

a) The pooled proportion here is computed as:
P = (64 + 76) / (307 + 179) = 0.2881

The standard error here is computed as:

$SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.2881(1 - 0.2881)(\frac{1}{307} + \frac{1}{179})} = 0.0426$

The sample proportions here are computed as:
p1 = 64/307 = 0.2085
p2 = 76/179 = 0.4246

Therefore the test statistic here is computed as:

$z^* = \frac{p_1 - p_2}{SE} = \frac{0.2085 - 0.4246}{0.0426} = -5.07$

therefore -5.07 is the test statistic value here.

As this is a two tailed test, we have from the standard normal tables, the p-value here:
p = 2P(Z < -5.07) = 0

Therefore 0 is the required p-value here.

b) For a true proportions of 0.24 and 0.42 and sample sizes as 307 and 179, and for 0.05 level of significance, we have from the standard normal tables:
P(-1.96 < Z < 1.96) = 0.95

Therefore the probabiltiy that the null hypothesis will be rejected is computed here as:

= 1 - Probability that null hypothesis is not rejected

The pooled proportion here is computed as:
P = (0.24*307 + 0.42*179) / (307 + 179) = 0.3063

The standard error here is computed as:

$SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.3063(1 - 0.3063)(\frac{1}{307} + \frac{1}{179})} = 0.0433$

The test statistic here is computed as:
Z = (p1 - p2) / SE = (0.24 - 0.42) / 0.0433 = -4.15

The probability now is computed here as:

$p_{diff} = - 0.18 \pm 1.96*0.0433$

$p_{diff} = - 0.18 \pm 1.96*0.0433 = -0.0951, -0.2649$

Therefore the probability here is computed as:

$= 1 - P(\frac{-0.2649 - 0}{0.0426} <Z < \frac{- 0.0951 -0}{0.0426})$

Getting it from the standard normal tables, we have here:

Therefore 0.9871 is the required probability here.