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: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. [tex] \bf H_0[/tex]: The life expectancy of people living in rural Idaho is 77 years [tex] \bf H_a[/tex]: 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 [tex] \bf t=\frac{\bar x-\mu}{s/\sqrt{n}}[/tex] where [tex] \bf \bar x[/tex] = 79.14 is the mean of the sample [tex] \bf \mu[/tex] = 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 [tex] \bf t=\frac{79.14-77}{2.48/\sqrt{11}}=2.862[/tex] Now, we obtain the critical upper value [tex] \bf t^*[/tex] 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 [tex] \bf t^*[/tex] such that the area under the t distribution to the left of [tex] \bf t^*[/tex] 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 [tex] \bf t^*[/tex] = 1.3722 Since our t-statistic is greater than [tex] \bf t^*[/tex] we can reject the null hypothesis and say this sample provide evidence that people living in rural Idaho communities live longer than 77 years.