(8.63) The state of Michigan tested 766 water samples from private wells in Lapeer county, Michigan and found that 112 exceeded the EPA guidelines of 50ppb. (a) Find the standard error of the estimated proportion. (b) Find the 95% confidence interval width for the true proportion. (c) Is normality an issue here? Explain (d) try the very quick rule. Does it work well here? Explain. Data is found from Detroit free press, November 27, 1997 p. 8A

x = 112, and n = 766

(a) Find the standard error of the estimated proportion.

Standard Error of the Sample Proportion:

(b) Find the 95% confidence interval width for the true proportion.

Half width = =1.960*0.0128 = 0.025

CI = (0.121, 0.171)

The 95% confidence interval width for the true proportion is 0.05 (=2*0.025).

(c) Is normality an issue here? Explain

The sample proportion p = x/n may be assumed normal if both n? ? 10 and n(1 ? ?) ? 10.

n? = 0.1462*766 = 112, and n(1 ? ?) = 766*(1 – 0.1462) = 654

Since, both n? ? 10 and n(1 ? ?) ? 10, therefore the normality is not an issue here.

(d) Try the very quick rule. Does it work well here? Explain.

Very Quick Rule

95% confidence interval: = (0.110, 0.182)

The very quick rule for 95% confidence interval does not work well here as it works well only if p is near 0.50 and in this case the value of p is 0.1462.

Confidence interval – proportion

95%

confidence level

0.146214099

proportion

766

n

1.960

z

0.025

half-width

0.171

upper confidence limit

0.121

lower confidence limit