Question

Would you favor spending more federal tax money on the arts? Of a random sample of n1 = 209 women, r1 = 57 responded yes. Another random sample of n2 = 194 men showed that r2 = 59 responded yes. Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts? Use α = 0.10. Solve the problem using both the critical region method and the P-value method. (Test the difference p1 − p2. Round the test statistic and critical value to two decimal places. Round the P-value to four decimal places.)

Answer 1

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the women and men. The population proportion of women and men who favor spending more federal tax dollars on the arts would be p1 and p2 respectively.

P1 - P2 = difference in the proportion of women and men who favor spending more federal tax dollars on the arts.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 ≠ p2

p1 - p2 ≠ 0

it is a two tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For women

x1 = 57

n1 = 209

P1 = 57/209 = 0.27

For men,

x2 = 59

n2 = 194

P2 = 59/194 = 0.3

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (57 + 59)/(209 + 194) = 0.29

1 - pc = 1 - 0.29 = 0.71

z = (P1 - P2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.27 - 0.3)/√(0.29)(0.71)(1/209 + 1/194) = - 0.03/0.045

z = - 0.67

Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.25

We would double this area to include the area in the right tail of z = 0.67 Thus

p = 0.25 × 2 = 0.5

By using the p value,

Since 0.1 < 0.5, we would accept the null hypothesis.

By using the critical region method,

The calculated test statistic is 0.67 for the right tail and - 0.67 for the left tail

Since α = 0.1, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.1/2 = 0.05

The z score for an area to the left of 0.05 is - 1.645

For the right, α/2 = 1 - 0.05 = 0.95

The z score for an area to the right of 0.95 is 1.645

In order to reject the null hypothesis, the test statistic must be smaller than - 1.645 or greater than 1.645

Since - 0.67 > - 1.645 and 0.67 < 1.645, we would fail to reject the null hypothesis.

Answer 2

Answer:

The null hypothesis failed to be rejected.

There is  not enough evidence to support the claim that the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts differs significantly.

Test statistic z = 0.70

P-value = 0.4871

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts differs significantly.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.10.

The sample 1, of size n1=209 has a proportion of p1=0.2727.

$p_1=X_1/n_1=57/209=0.2727$

The sample 2, of size n2=194 has a proportion of p2=0.3041.

$p_2=X_2/n_2=59/194=0.3041$

The difference between proportions is (p1-p2)=-0.0314.

$p_d=p_1-p_2=0.2727-0.3041=-0.0314$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{57+59}{209+194}=\dfrac{116}{403}=0.2878$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.2878*0.7122}{209}+\dfrac{0.2878*0.7122}{194}}\\\\\\s_{p1-p2}=\sqrt{0.00098+0.00106}=\sqrt{0.00204}=0.0451$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.0314-0}{0.0451}=\dfrac{-0.0314}{0.0451}=-0.696$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t$

As the P-value (0.4871) is bigger than the significance level (0.1), the effect is not significant.

The null hypothesis failed to be rejected.

There is  not enough evidence to support the claim that the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts differs significantly.