A simple random sample of 80 items resulted in a sample of 49. The population standard
deviation is σ = 7 .
i. What is the standard error of the mean? (1 Mark)
ii. At 99% probability, what is the margin of error?
i.
SE=σ/n=7/49=1SE=\sigma/\sqrt{n}=7/\sqrt{49}=1SE=σ/n=7/49=1
ii.
MOE=Z0.99σ2/n=2.576MOE=Z_{0.99}\sqrt{\sigma^2/n}=2.576MOE=Z0.99σ2/n=2.576
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