According to the latest survey the mean world lifespan is 68.8 years the university of Oregon wants to see if the average life span of the residence of Oregon differs from the world average suppose a study followed a random sample of 26 Oregonians and found their average lifespan to be 72.8 years with a standard deviation of 2.5 years

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.

50- 250 is correct i did that one too

step-by-step explanation:

answer: what are the answer choices?

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