Question 949:
Asked on April 29, 2012
Tags:
1-sample t-test
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Q949_1_q949.xls
We want to test the hypothesis that the percentage of silicon dioxide in cement is equal to 5. We will use a 1 sample t-test to find out. The Null Hypothesis is that the percent is equal to 5. The alternative hypothesis is that the percent is not equal to 5 and we will reject the Null if we have evidence at the .01 level.
- First we create the test statistic which is the difference between the sample mean and test mean : 5.21 - 5 = .21 divided by the standard error of the mean (SE).
- The standard error is equal to the sample standard deviation divided by the square root of the sample size = .38/SQRT(36) = .0633
- This makes the test statistics (t) = .21/.0633 = 3.32
- We now look up this test statistic using a t-table with 35 degrees of freedom. The attached excel file uses a formula to look it up for you. This generates a p-value of .002. [Note: You can also use the z instead of the t distribution but it's generally better to stick with the t for all sample sizes].
- Because the p-value of .002 is less than .01 we reject the Null Hypothesis and conclude there is evidence that the percent is different than 5%.