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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.


Sagot :

Answer:

The value of teh test statistic is [tex]z = 5.54[/tex]

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:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that [tex]\mu = 0.2, \sigma = \sqrt{0.2*0.8} = 0.4[/tex]

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

This means that [tex]X = 0.27, n = 1003[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.27 - 0.2}{\frac{0.4}{\sqrt{1003}}}[/tex]

[tex]z = 5.54[/tex]

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.