A random sample of 16 mid-sized cars tested for fuel consumption gave a mean of 26.4 kilometers per liter with a standard deviation of 2.3 kilometers per liter.
i) Assuming that the kilometers per liter given by all mid-sized cars have a normal distribution, find a 99% confidence interval for the population mean μ.
99% = mean ± 2.575 SEm
SEm = SD/√n
To calculate the 99% confidence interval for the population mean μ, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard deviation/sqrt(sample size))
1. Find the critical value:
Since we want a 99% confidence interval, we need to find the critical value that corresponds to a 99% confidence level. We can use a t-distribution because the sample size is small (n < 30) and the population standard deviation is unknown. To find the critical value, we need to determine the degrees of freedom, which is equal to the sample size minus 1 (n-1). In this case, the degrees of freedom is 16-1 = 15. Using a t-distribution table or calculator, we can find the critical value for a 99% confidence level and 15 degrees of freedom, which is approximately 2.947.
2. Calculate the confidence interval:
Now that we have the critical value, we can calculate the confidence interval using the formula mentioned earlier.
Confidence Interval = 26.4 ± (2.947 * 2.3/√16)
Calculating the expression inside the parentheses:
Standard deviation/sqrt(sample size) = 2.3/√16 = 2.3/4 = 0.575
Calculating the confidence interval:
Confidence Interval = 26.4 ± (2.947 * 0.575)
Confidence Interval = 26.4 ± 1.693
So, the 99% confidence interval for the population mean μ is (24.707, 28.093).