Sam thinks that there is a difference in quality of life between rural and urban

Question

Sam thinks that there is a difference in quality of life between rural and urban living. He collects information from obituaries in newspapers from urban and rural towns in Idaho to see if there is a difference in life expectancy. A sample of 11 people from rural towns give a life expectancy of x bar_r = 83.6 years with a standard deviation of s_r = 8.74 years. A sample of 8 people from larger towns give x bar_u = 72.3 years and s_u = 6.7 years. Does this provide evidence that people living in rural Idaho communities have different life expectancy than those in more urban communities? Use a 10% level of significance. (a) State the null and alternative hypotheses: (Type "mu _r" for the symbol mu _r, e.g. mu _r not = mu_ u for the means are not equal, mu _r > mu _ u for the rural mean is larger, mu _r

Explanation / Answer

a. State the hypotheses: Null hypothesis is the hypothesis of no difference. Therefore, it states that there is no difference in mean life expectancy of rural idaho and urban communities. It is of interest to know if there exists any difference in mean life expectancy of rural idaho and urban communities. The hypotheses are as follows:

H0: mu_r=mu_u

Ha: mu_r not= mu_u

b. The sample sizes for rural idaho and urban communities are different. Therefore, assume unequal population standard deviations.

Degrees of freedom is as follows:

df={(s1^2/n1+s2^2/n2)^2}/{1/n1-1(s1^2/n1)^2+1/n2-1(s2^2/n2)^2}, where, s denote sample standard deviation, n denotes sample size and 1 , 2 denote rural and urban sample.

={(8.74^2/11+6.70^2/8)}/{1/11-1(8.74^2/11)^2+1/8-1(6.70^2/8)^2}

=16

c. The test statistic is as follows:

t=(x1bar-x2bar)/sqrt(s1^2/n1+s2^2/n2), where, xbar denote sample mean.

=(83.60-72.30)/sqrt(8.74^2/11+6.70^2/8)

=3.19

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