Question

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.

Answer 1

The null hypothesis is that the mean is 83 out of 100.  The alternative hypothesis is that the mean is less than 83.

Answer 2

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

Explanation:

We are given the following in the question:

Population mean, μ = 83

Sample:

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

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(765)/(10) = 76.5`

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 8

First, we design the null and the alternate hypothesis

`H_(0): \mu = 83\\H_A: \mu < 83`

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

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(76.5 - 83)/((8)/(√(10)) ) = -2.57`

Now,
`z_(critical) \text{ at 0.05 level of significance } = -1.64`

Since,

`z_(stat) < z_(critical)`

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.