Question

To estimate the mean triglyceride levels of people in West Palm Beach, a blood test on 20 randomly selected people was conducted and yielded a mean triglyceride level of 200, with a standard deviation of 10. Construct a 95% confidence interval for the true average triglyceride level of people in West Palm Beach. Round your answer to three decimal places.
a. (218.868, 229.132)
b. (214.967, 233.033)
c. (219.741, 230.259)
d. (195.320, 204.680)

Answer 1

Answer:

lower limit = 195.320

upper limit =  204.680

so correct option is d. (195.320, 204.680)

Explanation:

given data

sample mean x = 200

standard deviation SD = 10

no of sample n = 20

confidence interval = 95%

solution

we get here first Z value by confidence interval i.e 95 %

so Z is by table is =  2.093

and we apply here margin of error formula that is express as

margin of error = Z × $\frac{SD}{\sqrt{n} }$    ....................1

put here value we get

margin of error = 2.093 × $\frac{10}{\sqrt{20} }$

margin of error =  4.6800

so here lower limit and upper limit will be as  

lower limit = sample mean - margin of error

lower limit = 200 - 4.6800

lower limit = 195.320

and  

upper limit = sample mean + margin of error

upper limit = 200 + 4.6800

upper limit =  204.680

so correct option is d. (195.320, 204.680)