Question

A professor finds that the average SAT score among all students attending his college is 1150 ± 150 (μ ± σ). He polls his class of 25 students and finds that the average SAT score is 1200. Suppose he computes a one-sample z test at a .05 level of significance. What is his decision

Answer 1

Answer:

Step-by-step explanation

Hello!

Be X: SAT scores of students attending college.

The population mean is μ= 1150 and the standard deviation σ= 150

The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.

If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:

H₀: μ = 1150

H₁: μ ≠ 1150

α: 0.05

$Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

$Z_{H_0}= \frac{1200-1150}{\frac{150}{\sqrt{25} } } = 1.67$

The p-value for this test is  0.0949

Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.

I hope it helps!