2 A pharmaceutical company wanted to estimate the population

2. A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 sales people. 35 sales people were randomly selected. Their mean monthly sales was $10,332 with a population standard deviation of $1,031. Construct a 95% confidence interval for the population mean.

3. A survey of households in a small town showed of 2,205 sampled households, that in 845 households at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

Solution

2.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    10332          
z(alpha/2) = critical z for the confidence interval =    1.959963985          

N = 250
n = 35

s = effective sample standard deviation =    sigma*sqrt[(N-n)/(N-1)] = 958.0280429          
n = sample size =    35          
              
Thus,              
Margin of Error E =    317.3893066          
Lower bound =    10014.61069          
Upper bound =    10649.38931          
              
Thus, the confidence interval is              
              
(   10014.61069   ,   10649.38931   ) [ANSWER]

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