This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STA 100: Homework 6 Due: 5:00pm, Wed 2nd March Assigned: Thursday, 24th February For all questions you must show your working. This enables us to understand your thought process, give partial credit and prevent crude cheating. Please see the code of the conduct in the Syllabus for rules about collaborating on homeworks. You must turn in a printed version of your R output with your solutions! (If you use another piece of software then that is fine please state this on your solutions) 1. Pooled vs. Unpooled Testing: Investigators are interested in the connection between lev els of calcium in the blood, and blood pressure. Data was collected for 97 patients, available on the course website. Each of the patients was classified as having either medium or high blood pressure, with the platelet calcium concentration (in nM) measured for each patient. (a) Make two boxplots of the data, one for each group. Note: You can do this in R using: Graphs &gt; Boxplot Then select by group . (b) Start by assuming the variance of calcium concentration is the same in each group. Perform a pooled twosample t test to test the hypothesis that calcium concentration is the same for both groups. Use a twosided test at level = 0 . 01. You must write out all of the following steps: The null and alternative hypotheses, The test statistic The reference distribution (i.e., the distribution of the test statistic under H ) 1 The p value for the test Your decision The interpretation in terms of the specific example Note: You can either do this by hand, or using the builtin function in R . To obtain the necessary quantities to do the test by hand: Statistics &gt; Summaries &gt; Numerical Summaries Then select Summarize by groups . To have R do the test for you: Statistics &gt; Means &gt; Independent Samples t test The null hypothesis : 1 = 2 , or, there is no difference in the levels of calcium between the two groups. The alternative hypothesis : 1 6 = 2 , or, the levels between the two groups are not equal. The test statistic is: t = x 1 x 2 s p q 1 n 1 + 1 n 2 = 3 . 0432 x 1 x 2 = 136 . 3318 118 . 8093 = 17 . 5225 n 1 = 36 ,n 2 = 61 s p 2 = ((36 . 07622 2 35 + 20 . 71717 2 60) / 95) = 750 . 57 s p = 27 . 40 The reference distribution is a t distribution with df equal to 95 ( n 1 + n 2 2). the p value is 0.003027. Given that the p value is less than , we should reject the null hypothesis. It means that there is connection between the levels of calcium in the blood and blood pressure....
View Full
Document
 Fall '10
 DRAKE

Click to edit the document details