Question

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.

Answer 1

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

Step-by-step 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 $\frac{ \alpha }{2}$ from the normal distribution table, the value is  

           $Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

           $E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{ \sqrt{n} }$

substituting values

            $E = 1.96 * \frac{ 4.86 }{ \sqrt{20 } }$

           $E = 2.13$

The  95% confidence interval is  mathematically represented as

          $\= x - E < \mu < \= x + E$

substituting values

         $100.4 - 2.13 < \mu < 100.4 + 2.13$

          $98.27 < \mu < 102.53$