You measure the lifetime of a random sample of 25 rats that are exposed to 10 sv of radiation (the equivalent of 1000 rem), for which the ld 100 is 14 days. the sample mean is = 13.8 days. suppose that the lifetimes for this level of exposure follow a normal distribution, with unknown mean μ and standard deviation σ = 0.75 day. what is the 95% confidence interval for μ based on these data?

CI 95%(μ)= [13.506 ; 14.094]

Step-by-step explanation:

The confidence interval (CI) formula is:

CI (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proportion of the distribution tails that are outside the confidence interval. In this case, 5% because 100-90%

Z(5%/2)= is the critical value of the standardized normal distribution. In this case is 1.96

σ= standard deviation. In this case 0.75 day

mean= 13.8 days

n= number of observations . In this case 25

Then, the confidence interval (90%) is:

CI 95%(μ)= 13.8+- [1.96*(0.75/sqrt(25)]

CI 95%(μ)= 13.8+- [1.96*(0.75/5) ]

CI 95%(μ)= 13.8+- (0.294)

CI 95%(μ)= [13.8-0.294 ; 13.8+0.294]

CI 95%(μ)= [13.506 ; 14.094]

were is the equation

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