6. A psychology professor of a large class became curious as to whether the students who turned in tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 students to turn in tests was 78. Using the .05 significance level, was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?

Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

Step-by-step explanation:

Given that;

U = 75

X = 78

standard deviation α = 10

sample size n = 20

population is normally distributed

PROBLEM is to test

H₀ : U = 75

H₁ : U ≠ 75

TEST STATISTIC

since we know the standard deviation

Z =  (X - U) / ( α /√n)

Z = ( 78 - 75 ) / ( 10 / √20)

Z = 1.3416 ≈ 1.342

Now suppose we need to test at ∝ = 0.05 level of significance,

Then Rejection region for the two tailed test is Zc = 1.96

∴ Reject H₀ if Z > Zc

and we know that Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

597

step-by-step explanation: