Chapter 7. Hypothesis Testing with One Sample

Chapter 7: Hypothesis Testing with One Sample 231Cumulative Review Exercises1. Monitoring DioxinHere is the data listed in order, and the necessary sums.0.018 0.0268 0.0281 0.032 0.044 0.0524 0.161 0.175 0.176Σx = 0.7133 Σx 2 = 0.09506a. The mean is x = Σx /n = 0.7133/ 9 = 0.07926b. The median is 0.044c. The standard deviation is s = nΣ(x 2 ) − (Σx) 29 ⋅ 0.09506− (0.7133)2= = 0.0694n(n −1)9 ⋅8d. The variance is s 2 = nΣ(x 2 ) − (Σx) 29 ⋅ 0.09506 − (0.7133)2= = 0.00482n(n −1)9 ⋅8e. The range is (0.176-0.018) = 0.158f. We assume that the data is from a normally distributed population. For a 95% confidence interval, α=.05,and so α/2=.025. Since n=9, the degrees of freedom df=n–1=8. Using Table A-3, we see that t .025= 2.306.The margin of error is E = t .025The confidence interval isx − E < µ < x + Esn= 2.3060.06949= 0.05330.07926 − 0.05335 < µ < 0.07926 + 0.053350.02591 < µ < 0.13261g. We again assume that the sample data come from a normally distributed population. The claim is that themean dioxin level is less than 0.16 ng/m 3 , so this is a left-tailed test. The sample size is n = 9 making thedegrees of freedom df = 8, the sample mean is x = 0.07926, and the sample standard deviation is s =0.0694. The significance level is 0.05.H 0 : µ = 0.16H 1 : µ < 0.16The test statistic is t = x − µ 0.07926− 0.16= =−3.490s / n 0.0694 / 9In a left-tailed test at the 0.05 significance level with df = 8, the critical value is −t α=−t .05=−1.860.In the row for 8 degrees of freedom, the absolute value for the test statistic is greater than 3.355 so the P-value is less than 0.005.We reject the null hypothesis.The sample data support the claim that the mean dioxin level is less than 0.16 ng/m 3 .h. One important characteristic of the data that is not addressed is that there was a dramatic drop in valuesover time, in that the 4 th and later measures were all much lower than the initial three measurements.