Question

Sue thinks that there is a difference in quality of life between rural and urban living. She 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 55 people from rural towns give a life expectancy of x¯r=72.3x¯r=72.3 years with a standard deviation of sr=9.77sr=9.77 years. A sample of 88 people from larger towns give x¯u=81.2x¯u=81.2 years and su=6.29su=6.29 years. Does this provide evidence that people living in rural Idaho communities have different average life expectancy than those in more urban communities? Use a 2% level of significanc

Answer 1

Answer:

Comparing the p value with the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and a would be a significant difference in the true mean lifetime expectancy number between rural and urban living.

Step by step explanation:

Data given and notation

$\bar X_{r}=72.3$ represent the mean for rural

$\bar X_{u}=81.2$ represent the mean for urban

$s_{r}=9.77$ represent the sample standard deviation for rural

$s_{u}=6.29$ represent the sample standard deviation for urban

$n_{r}=55$ sample size for the group rural

$n_{u}=88$ sample size for the group urban

t would represent the statistic (variable of interest)

$\alpha=0.02$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:$\mu_{u}=\mu_{r}$

Alternative hypothesis:$\mu_{u} \neq \mu_{r}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{u}-\bar X_{r}}{\sqrt{\frac{s^2_{u}}{n_{u}}+\frac{s^2_{r}}{n_{r}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value(s).

Based on the significance level$\alpha=0.02$ and $\alpha/2=0.01$ we can find the critical values from the t distribution dith degrees of freedom df=n1+n2-2=55+88-2=141, we are looking for values that accumulates 0.01 of the area on each tail on the t distribution.

For this case the two values are $t_{\alpha/2}=-2.35$ and $t_{1-\alpha/2}=2.35$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{81.2-72.3}{\sqrt{\frac{6.29^2}{88}+\frac{9.77^2}{55}}}}=6.02$

What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

$p_v =2*P(t_{141}>6.02)=1.43x10^{-8}$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and a would be a significant difference in the true mean lifetime expectancy number between rural and urban living.