4. A researcher is looking at the relative abundance of microbiota in the guts o
ID: 3330463 • Letter: 4
Question
Explanation / Answer
H0 : = 0.15
Ha : 0.15
Sample mean x = 0.1488
Sample standard deviation s = 0.052
standard error of the sample mean se0 = 0.052/ 5 = 0.0104
Test Statistic
t = (0.1488 - 0.15)/ (0.052/ sqrt(25) = -0.0012/ 0.0104 = 0.12
Reject Region :
95% CI
x +- t24,0.05 se0 = 0.1488 +- 2.064 * 0.0104 = (0.127, 0.170)
reject region : reject region is t > tcr which t > 2.064
p - value = 2 * Pr (t > 2.064) = 0.9055
we shalln't reject the null hypothesis and can conclude that population mean of microbiota relative abundance is not different from 0.15.
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