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In a random sample of 20 NBA basketball games the mean number of points scored by the home team was 100.4 with a standard deviation of 4.86.

Create and interpret a 95% confidence interval for the true mean number of points scored by an NBA basketball team at home.
You and your friend were watching a LA Lakers game where they were not playing at home. They only scored 98 points. Your friend says, "Wow, I bet if they were playing at home they would have scored a lot more points." Do you agree or disagree with your friend? Support your detailed answer.

User Alberta
by
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1 Answer

2 votes

Answer:

The 95% confidence interval is
98.27 < \mu < 102.53

This interval means that there 95% confidence that the true mean is within this interval

Yes i would agree with my friend because the lower and the upper limit 95% confidence interval for mean points scored at home is greater than 98 points

Explanation:

From the question we are told that

The sample size is n = 20

The sample mean is
\mu = 100.4

The standard deviation is
\sigma = 4.86

Given that the confidence level is 95% then the level of significance is mathematically evaluated as


\alpha = 100 - 95


\alpha = 5\%


\alpha = 0.05

Next we obtain the critical value of
( \alpha )/(2) from the normal distribution table, the value is


Z_{( \alpha )/(2) } = 1.96

Generally the margin of error is mathematically represented as


E = Z_{(\alpha )/(2) } * (\sigma)/( √(n) )

substituting values


E = 1.96 * ( 4.86 )/( √(20 ) )


E = 2.13

The 95% confidence interval is mathematically represented as


\= x - E < \mu < \= x + E

substituting values


100.4 - 2.13 < \mu < 100.4 + 2.13


98.27 < \mu < 102.53

User N Djel Okoye
by
5.5k points
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