Question

Andrew thinks that people living in a rural environment have a healthier lifestyle than other people. He believes the average lifespan in the USA is 77 years. A random sample of 11 obituaries from newspapers from rural towns in Idaho gives x¯=79.14 and s=2.48. Does this sample provide evidence that people living in rural Idaho communities live longer than 77 years? Use a 10% level of significance.

Answer 1

Answer:

Yes, it does.

Step-by-step explanation:

Since the data is approximately Normally distributed and the sample is less than 30, we will be using the Student's t Distribution.

$\bf H_0$: The life expectancy of people living in rural Idaho is 77 years

$\bf H_a$: The life expectancy of people living in rural Idaho is greater than 77 years.

So, this is a right-tailed test.

Our t-statistic is given by

$\bf t=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

$\bf \bar x$ = 79.14 is the mean of the sample

$\bf \mu$ = 77 is the mean of the null hypothesis

s = 2.48 is the sample standard deviation

n = 11 is the sample size

Computing our t-statistic we get

$\bf t=\frac{79.14-77}{2.48/\sqrt{11}}=2.862$

Now, we obtain the critical upper value $\bf t^*$ for a right-tailed test hypothesis corresponding to a 10% level of significance associated with the Student's t Distribution with 10 degrees of freedom (sample size-1). This is a value $\bf t^*$ such that the area under the t distribution to the left of $\bf t^*$ equals 10% = 0.01

We can do it either by looking up a table or a spreadsheet.

In Excel use

TINV(0.2,10)

In OpenOffice Calc use

TINV(0.2;10)

and we would get $\bf t^*$ = 1.3722

Since our t-statistic is greater than $\bf t^*$ we can reject the null hypothesis and say this sample provide evidence that people living in rural Idaho communities live longer than 77 years.