Question

The first eight quizzes in this class had average scores of 8,9,9,8,8,9,8,7. This gives a sample mean of 8.25 and a sample standard deviation of 0.661. Test the hypothesis that the true mean quiz score is 8.0 against the alternative that it is greater than 8.0. What is the t-test statistic for this test

Answer 1

Test statistic

`t = (x^(-) -mean)/((S)/(√(n) ) )`

t = 1.076

Explanation:

Step(i):-

Given Mean of the Population (μ) = 8.0

Mean of the sample (x⁻) = 8.25

Given data

8,9,9,8,8,9,8,7

Given sample size n= 8

Given sample standard deviation(S) = 0.661

Step(ii):-

Null hypothesis : H: (μ) = 8.0

Alternative Hypothesis :H:(μ) > 8.0

Degrees of freedom = n-1 = 8-1=7

Test statistic

`t = (x^(-) -mean)/((S)/(√(n) ) )`

`t = (8.25 -8.0)/((0.661)/(√(8) ) )`

t = 1.076

Critical value

t₍₇,₀.₀₅₎ = 2.3646

The calculated value t = 1.076 < 2.3646 at 0.05 level of significance

Null hypothesis is accepted

Test the hypothesis that the true mean quiz score is 8.0 against the alternative that it is not greater than 8.0