Question

50 POINTS

Answer 1

Answer:

95% confidence interval for the true average triglyceride level of people in West Palm Beach

(215.519 , 224.480)

Step-by-step explanation:

Step(l) :-

Given a blood test on 30 randomly selected people was conducted and yielded a mean triglyceride level of 220, with a standard deviation of 12.

Given sample size n = 30

mean of the sample x⁻ = 220

standard deviation of the sample  = 12

Step(ll):-

95% of confidence intervals:-

95% confidence interval for the true average triglyceride level of people in West Palm Beach

$(x^{-} - t_{0.95} \frac{S}{\sqrt{n} } , x^{-} + t_{0.95} \frac{S}{\sqrt{n} } )$

$(220 - t_{0.95} \frac{12}{\sqrt{30} } , 220+ t_{0.95} \frac{12}{\sqrt{30} } )$

The degrees of freedom ν = n-1 = 30-1 =29

t₀.₉₅ = 2.0452

$(220 - 2.0452 \frac{12}{\sqrt{30} } , 220+ 2.0452 \frac{12}{\sqrt{30} } )$

(220 - 4.4808 , 220 + 4.4808)

(215.519 , 224.480)

Conclusion:-

95% confidence interval for the true average triglyceride level of people in West Palm Beach

(215.519 , 224.480)