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In a pediatric clinic, a study is carried out to see how effective aspirin is in reducing temperature.

INSTRUCTIONS TO CANDIDATES
ANSWER ALL QUESTIONS

PART ONE: CONFIDENCE INTERVALS

In a pediatric clinic, a study is carried out to see how effective aspirin is in reducing temperature. Twelve 5-year-old children suffering from influenza had their temperatures taken immediately before and 1 hour after administration of aspirin. Suppose that we observe a mean difference in temperature was equal to 1.75 Fahrenheit and a standard deviation of differences equal to 0.8691 Fahrenheit.

[NOTE: Don’t round too soon in the calculations. Keep four decimal places as you work through the calculations and then round up to two decimal places for your final answer.]

Before calculating the 95% confidence interval, it is always a good plan to first identify the values for the key elements in the formula. From White (2020), we know that the formula for a confidence interval for a mean difference is:

 

where 

d ̅ = the sample’s estimate of the mean difference

t_((n-1)) = confidence factor

〖SD〗_d/√n = the sample’s estimate of the standard error of the difference

 

You will need to refer to Table A-3 from White (2020) to select the appropriate confidence factor for a 95%CI and a given degrees of freedom.  A copy of the table has been provided under the Biostats Assignment 2 link in Blackboard.

 

[NOTE: Always use a “0.05 area in two tails” in this class unless otherwise told.]

1) Based on the information provided in the Part One scenario, calculate the 95% confidence interval (95%CI) for the mean difference in temperature. Show you work and be sure to report both the lower and upper bounds of that interval.

 

 

 

2) Interpret the 95% CI.

 

 

 

3) As an added bonus, CIs can also be used to test a null hypothesis. In this scenario, we are told that the temperature levels of patients were measured before and 1 hour after of administration of aspirin. Let’s assume that the null hypothesis states that the mean difference in temperature level will be zero. Consider the 95%CI that you calculated in Question 1 above. Does the null value fall inside or outside of that 95% CI? Based on that, would you Reject or Fail to Reject the null hypothesis?

 

 

 

4) Calculate a 99% confidence interval (99%CI) for the mean difference in temperature. Show you work and report both the lower and upper bounds of that interval.

 

 

 

5) Interpret the 99% CI and describe how this 99%CI compares to the 95%CI that you calculated in Question 1.

 

 

 

PART TWO: PAIRED T-TEST

The degree of clinical agreement among different physicians on the presence or absence of generalized lymphadenopathy was assessed in 30 randomly selected participants from a prospective study of male sexual contacts of men with AIDS or an AIDS-related condition (ARC). For each participant, the total number of palpable lymph nodes was assessed twice – once by Doctor A and once by Doctor B. In this paired design, we want to examine whether there is a systematic difference between the assessments of Doctor A and Doctor B. Suppose that we observe the results in Table 1 (see the last page) from SAS when we analyzed our data using the UNIVARIATE procedure.

1) What is the purpose of a Paired t-Test and why is it the appropriate statistical test to conduct in this situation?

 

 

 

2) State the Null and Alternative hypotheses.

 

 

 

We can use information from the SAS output to calculate a 95%CI for the estimate of the 

mean difference. Towards the top of the table, we find the N of 30 and the mean difference of 2.733333. SAS has also calculated the standard deviation (2.87598) for us. We do need to determine the confidence factor for a sample size of 30; refer to Table A-3 provided in the assignment link.

3) Calculate and report the 95% CI. [Show your work.]

 

 

 

We can also use that information from the SAS output to calculate the test statistic (i.e. the

t-score). White (2020) note the t-score formula is:

 

where 

d ̅ = the sample’s estimate of the mean

0 = represents mean difference in the population based on the Null hypothesis

〖SD〗_d/√n = the sample’s estimate of the standard error of the difference

Note that the denominator in that equation is standard error (which SAS has already calculated for us).

4) Calculate the t-score using the values provided by SAS; show your work.

 

 

 

5) Statistical programs such as SAS do all the work for us. In question 4, you calculated the t-score test statistic. Take another look at the SAS output table – do you see that same test statistic value listed somewhere in the table?  If so, notice the reported P-value for that test statistic. Consider the supplemental video about using P-values to test hypotheses.

Based on the P-value approach, what conclusion would you make about the null hypothesis (i.e. would you Reject or Fail to Reject the null hypothesis)?

 

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