In your stor 155 class, the professor returns midterm 2 and claims that the average score was 83 out of 100 with a standard deviation of 8 points. when the professor steps out of the room, you quickly select a simple random sample of students (all students are present that day) and note their scores: 68, 75, 88, 79, 78, 79, 65, 77, 85, 71 a. you strongly suspect that the professor is overstating the average so that students will blame themselves if they have a low midterm score. state the null and alternative hypotheses for the appropriate test of significance at α=0.05.

We conclude that the professor overestimated the average score and the average score is less than 83.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 83

Sample:

68, 75, 88, 79, 78, 79, 65, 77, 85, 71

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 8

First, we design the null and the alternate hypothesis

We use One-tailed z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since,  

We reject the null hypothesis and accept the alternate hypothesis.

Thus, the professor overestimated the average score and the average score is less than 83.

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d