Answer:
Explanation:
We would set up the hypothesis test.
For the null hypothesis,
µ = 75
For the alternative hypothesis,
µ ≠ 75
Since the number of samples is 20 and no population standard deviation is given, the distribution is a student's t.
Since n = 20,
Degrees of freedom, df = n - 1 = 20 - 1 = 19
t = (x - µ)/(s/√n)
Where
x = sample mean = $69.46
µ = population mean = $75
s = samples standard deviation = $9.78
t = (69.46 - 75)/(9.78/√20) = - 2.53
We would determine the p value using the t test calculator. It becomes
p = 0.01
Since alpha, 0.05 > than the p value, 0.01, then the null hypothesis is rejected.
Therefore,
The value of the test statistic is -2.53; therefore, the null hypothesis is rejected for level of significance = 0.05