Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a 5% significance level. Test H0 : p=0.2 vs Ha : p≠0.2 using the sample results p^=0.27 with n=1003
Round your answer for the test statistic to two decimal places, and your answer for the p-value to three decimal places.

The value of teh test statistic is

The p-value of the test is 0 < 0.05, which means that there is significant evidence to conclude that the proportion differs from 0.2.

Step-by-step explanation:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that

Using the sample results p^=0.27 with n=1003

This means that

Value of the test statistic:

P-value of the test and decision:

The p-value of the test is the probability that the sample proportion differs from 0.2 by at least 0.07, which is P(|z| > 5.54), that is, 2 multiplied by the p-value of z = -5.54.

Looking at the z-table, z = -5.54 has a p-value of 0.

2*0 = 0.

The p-value of the test is 0 < 0.05, which means that there is significant evidence to conclude that the proportion differs from 0.2.

answer: $1794

step-by-step explanation:

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1