Question

This case study is about a study on coffee consumption of people that live in the United States. In total, 92 people were randomly selected and data was gathered regarding the annual consumption of coffee in gallons. Of these 92, 51 were females and 41 were males. You must show all work and formulas used in order to receive full credit. This includes showing all steps of the appropriate hypothesis test from the chapter notes. Round decimals in calculations to three places.

1. A consumer group claims that more than 62% of people choose coffee rather than other beverages as their preferred drink in the morning. In the sample of 92 people, 68 reported that they prefer coffee. At α = 0.02, is there enough evidence to support the consumer group’s claim? (12 points) Questions 2 and 3 refer to the following information regarding the annual consumption of coffee of the 41 males. Of the sample of the 41 males, the sample mean is 25.7 gallons and the sample standard deviation is 3.8 gallons. Assume the data is normally distributed and the sample is randomly selected.

2. Assuming α = 0.005, is there enough evidence to support a claim that the mean amount of consumption is at least 28 gallons? (12 points)

3. Assuming α = 0.05, is there enough evidence to support a claim that the standard deviation of the number of servings is 3 gallons? (12 points)

4. For the entire random sample of 92 people, the sample mean is 21.6 gallons. Assume the population standard deviation is 4.8 gallons and that the population is normally distributed. Is there enough evidence to support a claim that the mean amount of consumption is more than 20.2 gallons? Assume α = 0.03. (12 points)

5. Of the 92 people in the sample, the annual consumption of coffee and the weekly amount of hours worked were compared to find r = 0.623. Test the significance of this correlation coefficient using α = 0.10. Is there a significant correlation between annual coffee consumption and the weekly amount of hours worked? (12 points)

Answer 1

1:

Here we have -68, n 92, p- =0.739 X Hypotheses are Ho ps0.62 Н.:р>0.62 Level of significance: a 0.02 Test is right tailed. The standard deviation is: p(1-p) -sqrt((0.62*(1-0.62))/92) n 0.051 Test Statistics: р-р (0.739-0.62)/(0.051) Z = 2.333 Critical value: CV = 2.054 Rejection Region: If z > 2.054, Reject H0 Decision: Since test statistics lied in rejection region reject the null hypothesis. So we P-value: P-value-0.0098 reject the null hypothesis Decision: Since p-value is less than level of significance So we =NORMSINV(1-0.02) -1-NORMSDIST (2.333) Excel function for critical value: Excel function for p-value: b