Question

A laboratory tested 40 chicken eggs and found that the mean amount of cholesterol was 201 milligrams. It is known that the standard deviation of all such chicken eggs is s = 14.3 milligrams. Construct a 95 percent confidence interval for the true mean cholesterol content of all such eggs. O 197.3 <mu <204.7 196.6 < mu < 205.4 196.4 <mu <205.6 O 192.0 < < 210.0 Question 3 6 pts A randomly selected group of 40 bowlers from a city league had an average score was 192 with a population standard deviation of 8. Use this sample data to find a 95% confidence interval of the mean score u of all bowlers in this league. (188.5, 195.6) O (186.5, 197.5) (187.3, 196.1) (189.5, 194.5)

Answer 1

Solution :

Given that,

Solution :

Given that,

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 14.3 /  $\sqrt$40 )

= 4.4

At 95% confidence interval estimate of the population mean is,

     

T
- E < $\mu$ <
T
+ E

201 - 4.4 <  $\mu$ < 201 + 4.4

196.6 <  $\mu$ < 205.4

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 8 /  $\sqrt$40 )

= 2.5

At 95% confidence interval estimate of the population mean is,

     ± E

T

192 ±   192

(189.5 ,194.5)