PH1700HW6NLP

Nini Pham PH 1700 Homework 6 10.14: A 1980 study investigated the relationship between the use of OCs and the development of endometrial cancer [9]. The researchers found that of 117 endometrial- cancer patients, 6 had used the OC Oracon at some time in their lives, whereas 8 of the 395 controls had used this agent. Test for an association between the use of Oracon and the incidence of endometrial cancer, using a two-tailed test. ● Endometrial Cancer Patients Controls Total Used OC Oracon 6 8 14 Never used OC Oravon 111 387 498 Total 117 395 512 10.20: Otolaryngology Many children have tympanostomy tubes surgically inserted in their ears to reduce hearing loss associated with persistent otitis media and prevent recurrences of episodes of otitis media after tubes are inserted. However, acute otorrhea (a discharge from the external ear indicating inflammation of the external or middle ear), where middle ear fluid drains through the tube, is a common side effect with tympanostomy tubes. A clinical trial was conducted (Van Dongen et al. [10]) among children 1-10 years of age with prior symptoms of otorrhea comparing efficacy of (i) antibiotic eardrops, (ii) oral antibiotics, and (iii) observation without treatment, referred to below as observation. Children were seen at home by study physicians at 2 weeks and 6 months after randomization. The primary outcome was the presence of otorrhea at 2 weeks observed by study physicians. The results are given in Table 10.22. Provide a point estimate and a 95% CI for the prevalence of otorrhea in the ear drop group. ● ^ p = 4 76 = 0.052631 ● The point estimate for the prevalence of otorrhea in the ear drop group is 0.053.

Nini Pham ● ● We would use STATA to find the 95% confidence interval of the prevalence of otorrhea. Therefore, we are 95% confident that the prevalence of otorrhea in the ear drop group is between 1.45% and 12.93%. 10.21: What test can be used to compare the prevalence of otorrhea for the ear drop group vs. the observation group? State the hypotheses to be tested. ● Observed table: Eardrop Group Observation Group Total With Otorrhea 4 41 45 Without Otorrhea 72 34 106 Total 76 75 151 ● Expected table: Eardrop Group Observation Group Total With Otorrhea 45 × 76 151 = 22.65 45 × 75 151 = 22.35 45 Without Otorrhea 106 × 76 151 = 53.35 106 × 75 151 = 52.65 106 Total 76 75 151 ○ All the expected values are greater than 5, therefore, we will be using the chi- squared test. ○ H 0 : p 1 = p 2 ○ H a : p 1 ≠ p 2 ○ Null hypothesis: The cases and exposure variables are independent, therefore the prevalence of Otorrhea are independent of the treatment or observation group. ○ Alternate hypothesis: The cases and exposure variables are not independent, therefore the prevalence of Otorrhea is not independent of the treatment or

Nini Pham observation group. 10.22: Perform the test in Problem 10.21 and report a p-value (two-tailed). Interpret your results in words. ● X 2 =[ ( 22.65 − 4 ) 2 22.65 + ( 22.35 − 41 ) 2 22.35 + ( 53.35 − 72 ) 2 53.35 + ( 52.65 − 34 ) 2 52.65 ] ● X 2 = 15.36 + 15.56 + 6.52 + 6.61 ● X 2 = 44.05 ● The test statistic from the chi-square test is 44.05. ● df ( degreesof freedom )=( 2 − 1 ) × ( 2 − 1 )= 1 ● ● The p-value would be 3.2 × 10 11 ● The p-value is less than our significance level, 0.05, we would reject the null hypothesis and conclude the prevalence of otorrhea is different in the ear drop group compared to the treatment or observation. It is considered to be statistically significant that the cases and exposure variables are not independent. 10.38: Using the Hen data described in 10.38, what test procedure would be used to compare the percentage of hens whose pancreatic secretions were different when comparing saline to secretin. Perform and name the test requested. (Hint: the question can be reworded as: Assess if the proportion of hens whose pancreatic secretions increased was equal across the two different hormones. Or in other words, that hormone administered was independent of whether or not the secretions increased. Hint for data management: In order to avoid issues with multiple measures per hen, take the first measure of each hen for each hormone administered. State the null and alternative hypotheses, report the test statistic and p-value and interpret your result.) ● H 0 : p 1 = p 2 ● H a : p 1 ≠ p 2 ● Null hypothesis: Both the rows and columns variables are independent, therefore the increase in pancreas secretion is independent of the administered hormone. ● Alternate hypothesis: Both the rows and columns variables are not independent, therefore the increase in pancreas secretion is not independent of the administered hormone.