Answer to Question #348412 in Statistics and Probability for samuel

Question

 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.

Answer 1

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

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(2218-1.6449\times\dfrac{523}{\sqrt{38}},2218+1.6449\times\dfrac{523}{\sqrt{32}})$$ $$=(2078.4437, 2357.5563)$$

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