Previous research states, \"no evidence currently exists supporting or refuting

Question

Previous research states, "no evidence currently exists supporting or refuting the use of electric fans during heat waves" in terms of mortality and illness. Counterintuitively, Public Health guidelines suggest not using fans during hot weather, with some research reporting the potential of fans accelerating body heating. You decide to research further this seemingly contradictory guidance, hypothesizing that the true population proportion of heart and core temperature increases amidst higher ambient temperature and humidity levels is less than 38.3% and setting the level of significance at 5% for the formal hypothesis test. You randomly sample 18 participants based on your research funding and for 45 minutes, the study participants sit in a chamber maintained at a temperature of 108 degrees Fahrenheit (i.e., 42 degrees Celsius) and a relative humidity of 70%. At the end of the 45 minutes, you record for all participants if his/her heart and core temperature increased as compared to the start of the time period. The following table comprises the data you collect. Subject Heart and Core Temperature Increased? 1 0 2 0 3 0 4 0 5 1 6 0 7 1 8 0 9 0 10 1 11 1 12 1 13 1 14 0 15 0 16 0 17 1 18 0 Per Step 4 of the 5-Steps to Hypothesis Testing, compute the test statistic using the appropriate test statistic formula. Please note the following: 1) 0 and 1 are defined as no and yes, respectively, which is a typical coding scheme in Public Health; 2) you may copy and paste the data into Excel to facilitate analysis; and 3) do not round your numerical answer that you submit as the online grading system is designed to mark an answer correct if your response is within a given range. In other words, the system does not take into account rounding. On the other hand, rounding is preferable when formally reporting your statistical results to colleagues.

Explanation / Answer

n =18

yes = 7

no =11

h0= p =.38.3

h1= p<.38.3

significance level 5%

p = 7/18

q = 11/18

n =18

t =( p-.383)/(sqrt(p.q/n)

=0.005888889/0.1149044 =0.05125034

t critical at df 17 ~1.9

as t cricitcal > t calculated

we fail to reject null i.e p is not less that 38.3%

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