A random sample of size 16 is to be taken from a normal population having mean of 100 andvariance 4. What is the 90th percentile
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<u>Testing the hypothesis</u>, it is found that since the <u>p-value of the test is 0.0042 < 0.01</u>, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.
At the null hypothesis, it is tested if the <u>proportion is of 0.5</u>, that is:
At the alternative hypothesis, it is tested if the <u>proportion is different of 0.5</u>, that is:
The test statistic is given by:
In which:
is the sample proportion.
- p is the value tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are given by:
The value of the test statistic is:
Since we have a <u>two-tailed test</u>(test if the proportion is different of a value), the p-value of the test is P(|z| > 2.86), which is 2 multiplied by the p-value of z = -2.86.
Looking at the z-table, z = -2.86 has a p-value of 0.0021.
2(0.0021) = 0.0042
Since the <u>p-value of the test is 0.0042 < 0.01</u>, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.
A similar problem is given at brainly.com/question/24330815