13. You read that the average age of people living in the United States is 38.5. You believe that age is not
accurate for everyone living in Alabama. You randomly survey people living in Alabama and receive 41
responses. You calculate a sample mean of 37.8 years with a sample standard deviation of 1.5 years.
Perform a significance test for the mean using a significance level of a= 0.10 deviation of 1.5 years.
Perform a significance test for the mean using a significance level of and make a conclusion.
Null Hypothesis: The true mean age of people living in Alabama is equal to the national average of 38.5 years.
Alternative Hypothesis: The true mean age of people living in Alabama is not equal to the national average of 38.5 years.
We will use a two-tailed t-test since we don't have any specific directionality in our alternative hypothesis.
The test statistic is calculated as follows:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
t = (37.8 - 38.5) / (1.5 / sqrt(41))
t = -1.24
Using a t-distribution table with 40 degrees of freedom (sample size - 1), at a significance level of 0.10 and a two-tailed test, the critical values are -1.684 and 1.684.
Since our calculated t-value of -1.24 does not fall outside of the critical values, we fail to reject the null hypothesis.
Conclusion: Based on our significance test, we do not have sufficient evidence to conclude that the true mean age of people living in Alabama is significantly different from the national average of 38.5 years at a significance level of 0.10.