Question 949:

Asked on April 29, 2012

Tags: 1-sample t-test

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1. 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.

1. 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).
   
2. The standard error is equal to the sample standard deviation divided by the square root of the sample size = .38/SQRT(36) = .0633
3. This makes the test statistics (t) = .21/.0633 = 3.32
4. 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].
5. 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%.