Answer:
We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]
In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.
Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.
Therefore, We can select a lower confidence level and increase the sample size.
The best option that could be used to produce higher precision in our estimates of the population proportion is;
Option C; We can select a lower confidence level and increase the sample size.
Formula for confidence interval is given as;
CI = p^ ± z√(p^(1 - p^)/n)
Where;
p^ is the sample proportion
z is the critical value at given confidence level
n is the sample size
Now, the margin of error from the CI formula is:
MOE = z√(p^(1 - p^)/n)
Now, the lesser the margin of error, the narrower the confidence interval and thus the more precise is the estimate of the population proportion.
Now, looking at the formula for MOE, two things that could change aside the proportion is;
z and n.
Now, the possible values of z are;
At CL of 99%; z = 2.576
At CL of 95%; z = 1.96
At CL of 90%; z = 1.645
We can see that the higher the confidence level, the higher the critical value and Invariably the higher the MOE.
Thus, to have a narrow CI, we need to use a lower value of CL and increase the sample size.
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