A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4 . We
obtain a 95% confidence interval as ( 14.43 , 17.57 ) (14.43,17.57) . Which choice correctly interprets this interval? None of the answer options are correct. 95% of all lunches will cost between $14.43 and $17.57. 95% of the time, the average price for a lunch will be between $14.43 and $17.57. 95% of all samples of size n = 25 n=25 will have an average lunch price between $14.43 and $17.57.
A confidence interval of the mean is a range within is expected to find the population mean, with a certain degree of confidence.
In this case, the population mean is the average price for a lunch for a n=25 sample.
The interpretation of a 95% confidence interval of (14.43,17.57) should be that there is a 95% of confidence that the population mean (average price for a lunch out of a sample of 25 lunches) is within this interval.
the value of painting increases each year. To find the value of the painting for the next year, the art dealer multiplies the current value by 1.6. if the original value of the painting is 100, what is the value of the painting next yearStep-by-step explanation:
