Question 862:

Asked on April 12, 2010

1

Answer:

No answer provided yet.For the first question we use a 1-sample t-test to test the claim that the average age of college students is 29. We have a sample mean of 26, making an observed difference of 3 years.  The Null Hypothesis is that the mean age is 29.

To know if this difference is greater than chance fluctuations we divide it by the standard error of the mean (SE). The SE is found as the standard deviation divided by the square root of the sample size = 1.6/SQRT(24) = .326 getting us a test statistic t* of 3/.326 = 9.19.  We lookup this value in a t-table to get the probability of obtaining this large of a difference in standard errors at a sample size of 24 (23 degrees of freedom) and we get a p-value less than .00001.   Since our p-value is less than .01 we reject the Null Hypothesis and conclude the average age cannot be equal to 29.


For this second question we're given the population standard deviation so we can conduct a 1-sample Z test on the mean. The Null Hypothesis is that the average number of pieces of real fruit in a box is 50. We have a sample mean of 45 pieces making it an observed difference of 5 pieces. We divide this difference by the standard error (SE) which is the standard deviation divided by the square root of the sample size = 6/SQRT(100) = 6/10 = .6. Our test statistic z* = 5/.6 = 8.33. We lookup this value in a normal table and get a p-value less than .0001. We reject the Null Hypothesis and conclude the average number of pieces cannot be equal to 50.