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90%, 95% and 99% confidence intervals using normal distribution

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Mimi was the 5th seed in 2014 UMUC Tennis open that took place in August. In this tournament, she won 80 of her 100 serving games. Based on UMUC sports Network, she wins 75% of ;the serving games in her 5 year tennis career.

1. Find a 90 % confidence interval estimate of the proportion of serving games Mimi won. (Show work and round the answer to three decimal places).

2. The SAT scores are normally distributed. A simple random sample of 225 SAT scores has a sample mean of 1500 and a sample standard deviation of 300.

(a) What distribution will you use to determine the critical value? Why?

(b) Construct a 95% confidence interval estimate of the mean SAT score. (Show work and round the answer to tow decimal places).

(c ) Is a 99% confidence interval estimate of the mean SAT score wider than 95% confidence interval estimate you got from part (b)? Why? [You don't have to construct the 99% confidence interval]

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The solution gives detailed steps on determining the 90%, 95% and 99% confidence intervals using normal distribution in real example.

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1. Find a 90 % confidence interval estimate of the proportion of serving games Mimi won. (show work and round the answer to three decimal places)
At 90% level, critical value is 1.645 from standard normal table.
So a 90% confidence interval estimate of the proportion of serving games Mimi won = [0.80-1.645*sqrt[0.75*0.25/100], 0.80+1.645*sqrt[0.75*0.25/100]]=[0.729, 0.871]

2. ...

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