Suppose the coffee industry claimed that the average adult drinks 1.7 cups of co

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 50 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 0.5 cups. Using = 0.05, complete parts a and b below a. Is the coffee industry's claim supported A by this sample? Determine the null and alternative hypotheses The z-test statistic i:s (Round to two decimal places as needed.) The critical z-score(s) is(are) (Round to two decimal places as needed.

Explanation / Answer

Given :- sample mean is X¯=1.9 and the known population standard deviation is =0.5, and the sample size is n=50.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: =1.7

Ha: 1.7

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is =0.05, and the critical value for a two-tailed test is zc=1.96.

The rejection region for this two-tailed test is R={z:|z|>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

z =[(X¯–0) / (/n)] = [(1.9–1.7) / (0.5/50)] = 2.82

(4) Decision about the null hypothesis

Since it is observed that |z|=2.82 > zc=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0047, and since p=0.0047<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the coffee industry is supported by this sample.

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