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please answer the promt

21 Estii > Poll: Public Opir+ scussion_topics/1993 Learn by Doing Now take the opportunity to participate in this Learn by Do
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Answer #1

The random sample has 220 observations, Among them 22 has attended a women's basket ball game. The sample proportion , p=22/220=0.1

The standard error , S.E=\sqrt{p(1-p)/n}=0.02

M.E =S.E* z_{0.05}=0.02*1.96=0.0396

The Confidence interval is (p-M.E,p+M.E)=(0.06,0.14)

95% of the samples contain the true population proportion ie the proportion of all FootHill college students who have attended a women's basket ball game in other words the confidence interval constructed for 95% of the samples taken from the same population will have the true population proportion in it. ie There is good reason to believe that the population proportion lies between these two bounds of 0.06 and 0.14.since 95% of the time confidence intervals contain the true population proportion.


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please answer the promt 21 Estii > Poll: Public Opir+ scussion_topics/1993 Learn by Doing Now take the opportunity to participate in this Learn by Doing discussion. Prompt ons Recall that Foot...
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