Homework Sourse638 Refer to Exercise 637 In each case n 100 and n 400 con
6.38) Refer to Exercise 6.37. In each case (n = 100, and n = 400) construct a 95% confidence interval for the population proportion of mutants.
6.38ai) The 95% CI on p when n = 100 is: Lower bound = ? , upper bound = ?
Use 4th decimal accuracy.
6.38aii) The 95% CI on p when n = 400 is: Lower bound =? , upper bound = ?
Use 4th decimal accuracy.
Solution
6.37a)
As
sp (SE for sample proportion) = sqrt[ p (1 - p) / n ],
sp = sqrt(0.20*(1-0.20)/100) = 0.04 [ANSWER]
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b)
As
sp (SE for sample proportion) = sqrt[ p (1 - p) / n ],
sp = sqrt(0.20*(1-0.20)/400) = 0.02 [ANSWER]
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6.38a)
Note that
p^ = point estimate of the population proportion = x / n = 0.2
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.04
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
lower bound = p^ - z(alpha/2) * sp = 0.121601441
upper bound = p^ + z(alpha/2) * sp = 0.278398559
Thus, the confidence interval is
( 0.121601441 , 0.278398559 ) [ANSWER]
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6.38B)
Note that
p^ = point estimate of the population proportion = x / n = 0.2
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.02
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
lower bound = p^ - z(alpha/2) * sp = 0.16080072
upper bound = p^ + z(alpha/2) * sp = 0.23919928
Thus, the confidence interval is
( 0.16080072 , 0.23919928 ) [ANSWER]
