Here we have
"n =400"
"p_0=0.85"
The point estimate of population proportion is "\\hat p=\\frac{65}{400}=0.1625"
At 99% confidence level margin of error is
"ME=\\sqrt{\\frac{\\hat p(1-\\hat p)}{n}}\\cdot Z_{\\frac{1-0.99}{2}}"
"ME=\\sqrt{\\frac{0.1625\\cdot(1-0.1625)}{400}}\\cdot Z_{0.005}=\\sqrt{\\frac{0.1625\\cdot0.8375}{400}}\\cdot 2.58=0.0475"
"n\\cdot p_0= 400\\cdot 0.85 = 340 >10" and "n\\cdot (1 - p_0)= 400\\cdot (1-0.85) = 60 >10"
So, the condition is satisfied to use the large sample formula.
"\\hat p\\pm ME=0.1625\\pm 0.0475"
A 99% confidence interval for the population proportion is "(0.115, 0.210)"
Answer:
a) 0.1625
b) yes
c) 0.0475, (0.115, 0.210)
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