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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 :

Answer:

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

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