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a basketball scout randomly selected 144 players and timed howlong eah player took to perform a certain drill. the times in thissample were distributed with a mean of 8 minutes. the populationstandard deviation was known to be 3 minutes.
(a) find the critical value for a 95% confidence interval forthe population mean of drill times.
(b) using the 95% confidence interval,estimate the populationmean of drill times.

Formula for 95% confidence interval:

CI95 = mean ± 1.96 (sd/√n)
...where ± 1.96 represents the 95% interval using a z-table.

CI95 = 8 ± 1.96 (3/√144)

Finish the calculation.

I hope this helps.

no

We construct a new variable,

Z = (M - m) / [3 / sqrt(144)] = = 4 (M - m)

which then obeys the standard normal distribution.

We are looking for a value z such that P(-z < Z < z) = 0.95. To find this value using the table, we first use the symmetry of the standard normal distribution about its (zero) mean to write:

0.95 = P(-z < Z < z) = 2 P(0 < Z < z) = 2 [P(Z < z) - P(Z <= 0)] = 2 [P(Z < z) - 0.5] = 2 P(Z<z) - 1,

so P(Z < z) = 1.95 / 2 = 0.9750, which gives us the estimate z = 1.96.

Therefore, the confidence interval for M is [m - z/4, m + z/4] = [7.51, 8.49].

B. Not sure how to do the said "estimate." If we take the midpoint of the confidence interval as the estimate, we end up with M = m. A more pessimistic estimate would be closer to 7.51 than to 8.49.

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