Question

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm with a sample standard deviation of 1.9 cm. The heights of all bread loaves are assumed to be normally distributed. The baker is now interested in obtaining a 95% confidence interval for the true mean height of his loaves.

What is the lower bound to this confidence interval? cm (round to 2 decimal places)

What is the upper bound to this confidence interval? cm (round to 2 decimal places)

Answer 1

df = n -1 = 10 - 1 = 9

t critical value at 0.05 significance level with 9 df = 2.262

95% confidence interval for is

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

17 - 2.262 * 1.9 / sqrt(10) < < 17 + 2.262 * 1.9 / sqrt(10)

15.64 < < 18.36

Lower bound = 15.64

Upper bound = 18.36