1. Say that a supplier claims they are 99% confident that their products will be in the interval of 50.02 to 50.38. You take samples and find that the 99% confidence interval of what they are sending is 50.03 to 50.37. What conclusion can be made?

A) z distribution as the standard deviation always represents the population

B) Neither z nor t, as the population standard deviation is unknown with a small sample

C) t distribution as the sample standard deviation known with a small sample

D) t distribution as the data are assumed to be normally distributed

2. Under a time crunch, you only have time to take a sample of 15 water bottles and measure their contents. The sample had a mean of 20.05 ounces with a sample standard deviation of 0.3 ounces. What would be the 90% confidence interval, when we assumed these measurements are normally distributed?

A) (19.88, 20.22)

B) (19.92, 20.18)

C) (19.91, 20.19)

D) (19.75, 20.35)

3. Say that a supplier claims they are 99% confident that their products will be in the interval of 50.02 to 50.38. You take samples and find that the 99% confidence interval of what they are sending is 50.03 to 50.37. What conclusion can be made?

A) The supplier products have a higher mean than claimed

B) The supplier is less accurate than they claimed

C) The supplier is more accurate than they claimed

D) The supplier products have a lower mean than claimed