Answer:
Step-by-step explanation:
The standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean(μ) and standard deviation(σ)
The general formula for the sample size is given below:
[tex]n=p^{'}(1-p^{'})(\frac{Z_{\frac{a}{3} } }{E} )^{2}[/tex]
The formular for finding sample size is given as:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
a.)
it is given that [tex]E=±0.02, p^{'}_{1}=0.216, p^{'}_{2}=0.192[/tex]
The confidence level is 0.90
For (1 - ∝) = 0.90
∝=0.10; ∝/2 = 0.05
frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.216(1-0.216)+0.192(1-0.192))\\=351.22≅352[/tex]
The required sample size is 352 (nearest whole number)
b.)
it is given that [tex]E=±0.02, p^{'}_{1}=0.5, p^{'}_{2}=0.5[/tex]
The confidence level is 0.90
For (1 - ∝) = 0.90
∝=0.10; ∝/2 = 0.05
frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.5(1-0.5)+0.5(1-0.5))\\=541.205≅542[/tex]
The required sample size is 542 (nearest whole number)
