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Sagot :
Answer:
The p-value of the test is 0.1738, which means that for a level of significance above this, there is evidence of a change from the status quo.
Step-by-step explanation:
A high school has found over the years that out of all the students who are offered admission, the proportion who accept is 70%. Test if there is a change from status quo.
At the null hypothesis, we test if the proportion is 70%, that is:
[tex]H_0: p = 0.7[/tex]
At the alternate hypothesis, we test if there is a change from status quo, that is, the proportion is different from 70%. So
[tex]H_a: p \neq 0.7[/tex]
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.
70% is tested at the null hypothesis:
This means that [tex]\mu = 0.7, \sigma = \sqrt{0.7*0.3}[/tex]
Suppose they offer admission to 210 students and 156 accept.
This means that [tex]n = 210, X = \frac{156}{210} = 0.7429[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.7429 - 0.7}{\frac{\sqrt{0.7*0.3}}{\sqrt{210}}}[/tex]
[tex]z = 1.36[/tex]
P-value of the test and decision:
The p-value of the test is the probability that the sample proportion differs from 0.7 by at least 0.7429 - 0.7 = 0.0429, which is P(|z| > 1.36), which is 2 multiplied by the p-value of z = -1.36.
Looking at the z-table, z = -1.36 has a p-value of 0.0869
2*0.0869 = 0.1738
The p-value of the test is 0.1738, which means that for a level of significance above this, there is evidence of a change from the status quo.
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