Q1. Analyses of drinking water samples for 100 homes in each of two different se
ID: 3044873 • Letter: Q
Question
Q1. Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following means and standard deviations of lead levels (in parts per million): Section 1 Section 2 Sample size 100 100 4. 36.0 Standard deviation 5.96.0 Mean a. Calculate the test statistic to test for a difference in the two population means at the 5% level. Do the hypothesis testing by using critical value approach and and using p- value approach (by hand) b. Use a 95% confidence interval to estimate the difference in the mean lead levels for the two sections of the city. Does this interval confirm your conclusions in part a?Explanation / Answer
a)
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
n1 = 100 , x1 = 34.1 , s1= 5.9
n2 = 100 , x2 = 36 , s2 = 6
SE = sqrt[(s1^2/n1) + (s2^2/n2)]
SE = sqrt[(5.9^2/100) + (6^2/100)]
SE = 0.841
t = [ (x1 - x2) - d ] / SE
= [ (34.1 - 36) - 0 ] /0.84
= -2.259
Critical value at 5% = 1.984
p value at 5% = 0.026
We reject the null hypothesis
b)
z value at 95% = 1.96
CI = ( x1 - x2) +/- z * sqrt[(s1^2/n1) + (s2^2/n2)]
= (34.1 - 36) + /- 1.96 * sqrt[(5.9^2/100) + (6^2/100)]
= (-3.548 ,-0.255)
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