Answer:
(0.6430, 0.8170)
Step-by-step explanation:
Given that during a recent drought a water utility in a certain town sampled 100 residential water bills and found out that 73 of the residences had reduced their water consumption over that of the previous year.
Sample size n = 100
Sample proportion p = [tex]\frac{73}{100} =0.73[/tex]
q = 1-p = 0.23
Std error of proportion = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.73*0.27}{100} } \\=0.04440[/tex]
95% Z critical value = 1.96
Margin of error = [tex]1.96*0.0444 = 0.0870[/tex]
Confidence interval = sample proportion ±margin of error
0.642983946
0.817016054
(0.6430, 0.8170)
The 95% confidence interval for the proportion of residences that reduced their water is (0.6430, 0.8170).
This is used to determine the measure of how much uncertainty there is with any particular statistic.
Sample size (n) = 100
Sample proportion(p) = 73/100 = 0.73
q = 1-p = 0.23
√pq/n = √0.73×0.27/100 = 0.04440
95% Z critical value = 1.96
Margin of error = 1.96×0.04440 = 0.0870
Confidence interval = sample proportion ± margin of error
= (0.6430, 0.8170)
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