Answer:
Step-by-step explanation:
We would set up the hypothesis test.
For the null hypothesis,
µ = 68.8
For the alternative hypothesis,
µ ≠ 68.8
This is a two tailed test.
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 26,
Degrees of freedom, df = n - 1 = 26 - 1 = 25
t = (x - µ)/(s/√n)
Where
x = sample mean = 72.8
µ = population mean = 68.8
s = samples standard deviation = 2.5
t = (72.8 - 68.8)/(2.5/√26) = 8.16
We would determine the p value using the t test calculator. It becomes
p < 0.00001
Assuming a significance level of 0.05, then
Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the average life span of the residence of Oregon differs from the world average.
