Can someone help me on B and C
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Answer:
We conclude that the mean IQ of her students is different from the national average.
Step-by-step explanation:
We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.
The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113
Let, Null Hypothesis, :
= 100 {means that the mean IQ of her students is same as of national average}
Alternate Hypothesis, :
100 {means that the mean IQ of her students is different from the national average}
(a) The test statistics that will be used here is One sample z-test statistics;
T.S. = ~ N(0,1)
where, Xbar = sample mean score = 113
s = population standard deviation = 15
n = sample of students = 144
So, test statistics =
= 10.4
Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the mean IQ of her students is different from the national average.