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sdas [7]
3 years ago
15

According to the latest survey the mean world lifespan is 68.8 years the university of Oregon wants to see if the average life s

pan 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
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

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.

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