Question

The sample proportion of Americans who are dissatisfied with education in kindergarten through grade 12 is 0.55 with sample of size 1000. We are interested in testing at the 5% significance level, if this sample provides evidence that the proportion of Americans who are dissatisfied with education in kindergarten through grade 12 differs significantly from 50%.

Required:
What is the p-value?

Answer 1

The p-value for this hypothesis test is calculated using the z-test for proportions formula. If the resulting p-value is less than the significance level of 0.05, the null hypothesis that the proportion is 0.50 is rejected.

Step-by-step explanation:

To determine the p-value for the hypothesis test that checks if the proportion of Americans dissatisfied with education in kindergarten through grade 12 differs significantly from 50%, the following calculation is made using a z-test for proportions.

The sample proportion is 0.55 with a sample size of 1000, and we are testing against the null hypothesis that the true proportion (p) is 0.50.

The formula for the z-score in a test of proportions is:

Z = (p' - p) / sqrt[p(1 - p)/n]

where p' is the sample proportion, p is the hypothesized proportion, and n is the sample size. Plugging the values,

Z = (0.55 - 0.50) / sqrt[0.50 * (1 - 0.50) / 1000]

Z = 0.05 / sqrt[0.50 * 0.50 / 1000]

Z = 0.05 / 0.0158 = 3.16

Since this is a two-tailed test, we look for the probability of the z-score being greater than 3.16 or less than -3.16. Consulting the z-table or using a calculator with the normal distribution function tells us the p-value for each tail. We then double the p-value for the single tail to account for both tails in the two-tailed test.

Thus, if the calculated p-value is less than the alpha level of 0.05, we reject the null hypothesis, concluding there is sufficient evidence that the proportion differs from 50%.

Answer 2

p-value = 0,000755

We reject H₀. We can support that Americans who are dissatisfied with education in kindergarten throug grade 12 differs significantly from 50%

Step-by-step explanation:

Sample proportion : p₁ = 0,55

Sample size n = 1000

q₁ = 1 - p₁ q₁ = 1 - 0,55 q₁ = 0,45

Hypothesis Test:

Null Hypothesis H₀ p₁ = 0,5 ( 50 % )

Alternative Hypothesis Hₐ p₁ ≠ 0,5

Alternative hypothesis indicates that we have to develop a two-tail test

as n = 1000

p₁*n = 0,55 * 1000 = 550 q₁*n = 0,45*1000 = 450

Both products are bigger than 5. We can use the approximation of binomial distribution to normal distribution.

Significance level α = 5% α = 0,05 and α/2 = 0,025

z(s) = ( p₁ - 0,5 ) / √p₁*q₁/1000

z(s) = 0,05 * 31,62 / 0,4974

z(s) = 3,1785

From z-table we find p-value = 0,000755

Therefore p-value < 0,025

We are in the rejection region, we reject H₀