Question

Whiting, Indiana, leads the "Top 100 Cities with the Oldest Houses" list with the average age of houses being 66.4 years. Farther down the list resides Franklin, Pennsylvania, with an average house age of 59.4 years. Researchers selected a random sample of 24 houses from Whiting and 21 houses from Franklin and obtained the following statistics. Ata 0.05, can it be concluded that the houses in Whiting are older? Use theP-value method. Assume the variables are normally distributed and the variances are unequal. Mean age(in years) Standard deviation(in years) Whiting 62.9 5.4 Franklin 58.1 3.9

Answer 1

Given:

Sample size for Whiting, n1 =24

Sample size for Franklin, n2 =21

Mean age in Whiting =$\bar{X1}$ =62.9

Std.deviation in Whiting =S1 =5.4

Mean age in Franklin =$\bar{X2}$ =58.1

Std.deviation in Franklin =S2 =3.9

Welch's t-test or unequal variances t-test:

Null Hypothesis (H0): $\mu1 \leq \mu2$​​​​​​

Alternative Hypothesis (H1): $\mu1 >\mu2$​​​​​​ (right tailed test)

(where $\mu1$ =Population mean age in Whiting and $\mu2$ =Population mean age in Franklin).

Test statistic, t=$(\bar{X1} - \bar{X2})/\sqrt{(S1^2/n1)+ (S2^2/n2)}$ =$(62.9 - 58.1)/\sqrt{(5.4^2/24)+ (3.9^2/21)}$ =3.447

v1=n1-1 =23 and v2 =n2-2 =20

Degrees of freedom, V =$[S1^2/n1)+(S2^2/n2)]^2/[(S1^4/n1^2v1)+(S2^4/n2^2v2)]$ =$[5.4^2/24)+(3.9^2/21)]^2/[(5.4^4/24^2*23)+(3.9^4/21^2*20)]$ =42

From Student's T distribution table, at V =42 and $\alpha =0.05$​​​​​​ for right tailed test, t_critical =1.682​​​​​​

Conclusion: t > t_critical ​​​​​​ (3.447 > 1.682). There is a sufficient evidence to reject the null hypothesis at 0.05 significance level. Hence, there is a statistically significant evidence to claim that the mean age of houses in Whiting is greater than the mean age of houses in Franklin ($\mu1 > \mu2$​​​​​​).

Thus, it can be concluded that the houses in Whiting are older.