Question

The performance of a coffee vending machine is under inspection. Here are the reported amount of coffee for 7 cups : 8.06, 7.69, 7.36, 8.42, 8.33, 7.44, 7.20 oz. Find a 99% confidence interval of this machine’s target amount of coffee per cup. You may assume normal distribution when it is necessary.

Answer 1

From the given data

= Xi / n = 7.7857

S = Sqrt( Xi² - n ² )

= 0.4876

99% confidence interval for is

- t * S / sqrt( n) < < + t * S / sqrt( n)

7.7857 - 3.707 * 0.4876 / sqrt(7) < < 7.7857 + 3.707 * 0.4876 / sqrt(7)

7.1025 < < 8.4689

99% confidence interval is (7.1025 , 8.4689)