Question #348412

 Suppose a random sample of 38 sports cars has an average annual fuel cost of K2218 and the standard deviation was K523. Construct a 90% confidence interval for μ. Assume the annual fuel costs are normally distributed.


1
Expert's answer
2022-06-06T13:34:09-0400

Based on the information provided, the significance level is α=0.10,\alpha = 0.10,  and the critical value for a two-tailed test is zc=1.96.z_c =1.96.

The corresponding confidence interval is computed as shown below:



CI=(xˉzc×σn,xˉzc×σn)CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}})=(22181.6449×52338,2218+1.6449×52332)=(2218-1.6449\times\dfrac{523}{\sqrt{38}},2218+1.6449\times\dfrac{523}{\sqrt{32}})=(2078.4437,2357.5563)=(2078.4437, 2357.5563)

Therefore, based on the data provided, the 90% confidence interval for the population mean is 2078.4437<μ<2357.5563,2078.4437 < \mu < 2357.5563, which indicates that we are 90% confident that the true population mean μ\mu  is contained by the interval (2078.4437,2357.5563).(2078.4437, 2357.5563).

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