Respuesta :
Answer:
Option B) Reject the null hypothesis. There is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.
Test statistics = 2.036
P- value = 2.07% .
Step-by-step explanation:
We are given that the population mean, [tex]\mu[/tex] = 550 {National mean SAT score}
We have to test the principal claim that the mean SAT score in math at his school is better than the national mean score or not.
Let Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 550
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 550
The test statistics we will use here is;
[tex]\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows N(0,1)
where, Xbar = 574 and [tex]\sigma[/tex](Population standard deviation) = 100
n = sample of students = 72
Test Statistics = [tex]\frac{574 - 550}{\frac{100}{\sqrt{72} } }[/tex] = 2.036
Now at 5% level of significance the z table gives the critical value of 1.6449 and our test statistics is more than this as 2.036 > 1.6449 so we have sufficient evidence to reject null hypothesis and conclude that there is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.
P - value = P(Z > 2.04) = 0.0207 or 2.07%
Here also our P- value is less than the level of significance of 5% ,we will reject null hypothesis.
