A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.

Respuesta :

Answer:

[tex]CI=189.5,194.5[/tex]

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=40[/tex]

Mean [tex]\=x =192[/tex]

Standard deviation[tex]\sigma=8[/tex]

Significance Level [tex]\alpha=0.05[/tex]

From table

Critical Value of [tex]Z=1.96[/tex]

Generally the equation for momentum is mathematically given by

 [tex]CI =\=x \pm z_(a/2) \frac{\sigma}{\sqrt{n}}[/tex]

 [tex]CI =192 \pm 1.96 \frac{8}{\sqrt{40}}[/tex]

 [tex]CI=192 \pm 2.479[/tex]

 [tex]CI=189.5,194.5[/tex]