A basketball coach wants to know how many free throws an NBA

A basketball coach wants to know how many free throws an NBA player shoots during the course of an average practice. The coach takes a random sample of 43 players and finds the average number of free throws shot per practice was 225 with a standard deviation of 35. Construct a 99% confidence interval for the average number of free throws in practice.

[210.5995, 239.4005]__

[214.2290, 235.7710]__

[211.2506, 238.7494]__

[210.6155, 239,3845]__

We draw a random sample of size 36 from a population with standard deviation 3.5. If the sample mean is 27, what is a 95% confidence interval for the population mean?

[25.8567, 28.1433]__

[26.0405, 27.9595]__

[26.8100, 27.1900]__

[26.8401, 27.1599]__

Solution

1.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    225          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    35          
n = sample size =    43          
              
Thus,              
Margin of Error E =    13.74836002          
Lower bound =    211.25164          
Upper bound =    238.74836          
              
Thus, the confidence interval is              
              
(   211.25164   ,   238.74836   )
[The closest is OPTION C: [211.2506, 238.7494]. Maybe they used a table and that produces a little roundoff error.]

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