Question

1. A vending machine that dispenses coffee into cups must fill the cups with 7.8 oz of liquid. Before selling the vending machine to a college or business, the company tests the machine to be sure it is dispensing an average amount of 7.8 oz of coffee. A sample of 20 amounts is shown here. Assume that the amount of coffee dispensed is normally distributed. 7.78 7.79 7.82 7.82 7.87 7.84 Ounces Dispensed 7.80 7.80 7.83 7.85 7.82 7.78 7.75 7.83 7.84 7.73 7.82 7.87 7.81 7.88 (a) Set up the null and alternative hypotheses to test whether average amount of coffee dispensed is different from 7.8 ounces. (b) Test the hypotheses from part (a) with a = 0.05 by computing a critical region. Is there enough evidence to reject the null hypothesis? (c) Compute the p-value corresponding the hypotheses testing in part (a). What is your conclusion when a = 0.05? Note: yes, this part of the question is very similar to the previous part of the question. (d) Now use a = 0.1 for part (c). What is your decision?

Answer 1

1:

First we need to find the mean and SD of the sample data. Following table shows the calculations:

	X	(X-mean)^2
	7.78	0.0016
	7.82	0
	7.87	0.0025
	7.8	0.0004
	7.8	0.0004
	7.83	1E-04
	7.85	0.0009
	7.84	0.0004
	7.82	0
	7.81	0.0001
	7.79	0.0009
	7.82	0
	7.84	0.0004
	7.82	0
	7.78	0.0016
	7.75	0.0049
	7.83	1E-04
	7.73	0.0081
	7.87	0.0025
	7.88	0.0036
Total	156.33	0.0285

Sample size n =20 Sample sum: x-156.33 Sample mean: =156.33/20 = =7.82 Sample standard deviation: =sqrt(0.0285/(20-1)) n-1 -0.0387

(a-c)

Hypotheses are: Н : и3 7.8 На: и+ 7.8 Level of significance: a 0.05 Test is two tailed. Degree of freedom: df =n-1-19 Test Statistics: (7.82-7.8)/(0.0387/sqrt(20)) s/n 2.311 Critical value: Rejection Region 2.093 If It >2.093, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value 0.03 P-value: Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: =TINV(0.05,19) -TDIST (2.311,19,2) Excel function for p-value:

(d)

Since p-value is less than 0.10 so we reject the null hypothesis.