Suppose the coffee industry claimed that the average adult drinks 1.7 cups of coffee per day. To test this claim, a random sample of 40 adults was selected, and their average coffee consumption was found to be 1.9 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.6 cups. Using alphaequals0.10, complete parts a and b below. a. Is the coffee industry's claim supported by this sample? Determine the null and alternative hypotheses.

(a) No, the coffee industry's claim is not supported by this sample.

(b) Null hypothesis: The average adult drinks 1.7 cups of coffee per day.

Alternate hypothesis: The average adult drinks more than 1.7 cups of coffee per day.

Step-by-step explanation:

(a) Test statistic (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 1.9 cups per day

population mean = 1.7 cups per day

sd = 0.6 cups per day

n = 40

z = (1.9 - 1.7) ÷ 0.6/√40 = 0.2 ÷ 0.095 = 2.11

The test is a one-tailed test. Using alpha (significance level) = 0.1, the critical value is 2.326.

Conclusion:

Reject the null hypothesis because the test statistic 2.11 falls within the rejection region of the critical value 2.326.

The coffee industry's claim is contained in the null hypothesis, hence it is not supported by the sample because the null hypothesis is rejected.

(b) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

f(x) = y

f(-3) = -5

step-by-step explanation: