Question

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 60 adults was​ selected, and their average coffee consumption was found to be 1.85 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.4 cups. Using alpha α=0.10​, complete parts a and b below.

a. Is the coffee​ industry's claim supported by this​ sample?

Determine the null and alternative hypotheses.

H0​: μ ▼ less than < not equals ≠ greater than or equals ≥ equals = less than or equals ≤ greater than >

H1​: μ ▼ greater than > equals = greater than or equals ≥ less than < not equals ≠ less than or equals ≤

\The​ z-test statistic is What? . ​(Round to two decimal places as​ needed.)

The critical​ z-score(s) is(are) What? . ​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Because the test statistic ▼ (does not fall within the critical values, is greater than the critical value, falls within the critical values, is less than the critical value), pick answere ▼ (reject, do not reject) pick answer the null hypothesis.

b. Determine the​ p-value for this test. The​ p-value is what?. ​(Round to three decimal places as​ needed.)

Answer 1

Given that the average adult drinks = 1.7 cups of coffee per day. To test this​ claim, a random sample of n = 60 adults was​ selected, and their average coffee consumption was found to be = 1.85 cups per day.

the standard deviation of daily coffee consumption per day is = 0.4 cups.

a) So, based on the claim the hypotheses are:

Based on the Hypothesis it will be a two-tailed test.

Test Statistic:

The test statistic is calculated as:

Z = 2.91

The critical​ z-score(s):

The critical scores at 0.10 level of significance for a normal distribution are calculated using excel formula for normal distribution which is =NORM.S.INV(0.95), so the Critical values calculated are +1.645, -1.645

So, reject the Ho if Z is greater than 1.65 or Z is less than -1.65

Decision:

Because the test statistic does not fall within the critical values, so, we reject the null hypothesis.

b) P-value:

The P-value is also computed using the excel formula for normal distribution at calculated Z score, the formula used is

=2*(1-NORM.S.DIST(2.91, TRUE)).

The P-value resulted as 0.004.