Salaries of 31 college graduates who took
a course in college have a mean, x of
66,700 a standard deviation , o of 17,120, construct a 90% confidence interval for estimating the population
mean u.
66700 -1.645* 17120/sqrt((31)) =
66700 +1.645* 17120/sqrt((31)) =?
( , )?
To construct a confidence interval for estimating the population mean, u, with a given level of confidence, you can follow these steps:
Step 1: Determine the critical value. Since we want to construct a 90% confidence interval, we need to find the corresponding critical value from the t-distribution table. With 31 degrees of freedom (n - 1), the critical value can be found to be approximately 1.697.
Step 2: Calculate the standard error. The standard error (SE) is a measure of how much the sample mean, x, is likely to vary from the true population mean, u. It is calculated by dividing the standard deviation, o, by the square root of the sample size, which in this case is 31. Therefore, SE = o / sqrt(n) = 17020 / sqrt(31).
Step 3: Calculate the margin of error. The margin of error (E) represents the maximum likely difference between the sample mean, x, and the true population mean, u. It is calculated by multiplying the critical value by the standard error. E = critical value * SE = 1.697 * (17020 / sqrt(31)).
Step 4: Calculate the lower and upper bounds of the confidence interval. The lower bound is obtained by subtracting the margin of error from the sample mean (x), while the upper bound is obtained by adding the margin of error to the sample mean. Lower bound = x - E and Upper bound = x + E.
Now, let's calculate the confidence interval:
Lower bound = 66700 - (1.697 * (17020 / sqrt(31)))
Upper bound = 66700 + (1.697 * (17020 / sqrt(31)))
Plug in these values into the equations and calculate the confidence interval for estimating the population mean, u, at a 90% confidence level.