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State the null and alternate hypotheses and find the degree of freedom.

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 11 people from rural towns give a life expectancy of xr¯=68.4 years with a standard deviation of sr=7.04 years. A sample of 10 people from larger towns give xu¯=74.4 years and su=6.23 years. Does this provide evidence that people living in rural Idaho communities have different life expectancy than those in more urban communities? Use a 5% level of significance.

(a) State the null and alternative hypotheses: (Type ‘‘mu_r″ for the symbol μr , e.g. mu_rnot=mu_u for the means are not equal, mu_r>mu_u for the rural mean is larger, mu_r<mu_u , for the rural mean is smaller. )
H0 = ?

Ha = ?

(b) The degree of freedom is ?

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Walter B. | Success-Based Tutor Specializing in Your StudentSuccess-Based Tutor Specializing in Your...
4.9 4.9 (532 lesson ratings) (532)
Null hypothesis: μrural =  μurban
Alternative hypothesis: μrural ≠ μurban
Variance = s12/n1 + s22/n2 = 7.042/11 + 6.232/10 = 8.38689
test statistic = (74.4 - 68.4)/sqrt(8.38689) = 6/2.89601 = 2.072
using a calculator, the p value is calculated as 5.2%
since the p value is greater than 5%, the null hypothesis is not rejected
Andy C. | Math/Physics TutorMath/Physics Tutor
4.9 4.9 (19 lesson ratings) (19)
Null hypothesis: the means are different ---> their sum is NOT zero
Alternative hypothesis: the means are the same ---> their sum is zero

The formulas are found at:

Despite the variance are reasonably close, I chose not
to use the pooled test statistic just to play it safe:

X1 = 68.4
s1 = 7.04

X2 = 74.4
S2 = 6.23

degrees of freedom
s1^2/n1 + s2^2/n2 =8.034045454545....

C = 0.5608133567941....

The degrees of freedom is 100 / {10*C^2 + (1-c)^2*10}=19.708451741 which is rounded DOWN to 19.

The test statistic is (74.4 - 68.4)/sqrt(8.03045454545454....) = 2.117294095...

Per the T-Distribution table since the sample size is LESS than 30....
For a 1 tailed test you are 95% certain the means are different.
For a 2 tailed test you are 97.5% certain the means are different.


Null hypotheses always contain the equal sign. Therefore the null hypothesis is that the two means are equal and the alternative hypothesis is that the two means are different.
Next, not sure how you calculated the standard deviation for the combined sample. Once again, happy to send you the formula for pooled and un-pooled variance because your answers are incorrect.
Finally, your conclusions are incorrect. We do not have enough evidence to reject the null.