Question

The R.D. Wilson Company makes a soft drink dispensing machine that allows customers to get soft drinks from the machine in a cup with ice. When the machine is running properly, the average number of fluid ounces in the cup should be 14. Periodically the machines need to be tested to make sure that they have not gone out of adjustment. To do this, six cups are filled by the machine and a technician carefully measures the volume in each cup. In one such test, the following data were observed:_______.

14.25 13.7 14.02
14.13 13.99 14.0
Based on these sample data,which of the following is true if the significance level is .05?
A) No conclusion can be reached about the status of the machine based on a sample size of only six cups.
B) The null hypothesis cannot be rejected since the test statistic is approximately t = .29,which is not in the rejection region.
C) The null hypothesis can be rejected since the sample mean is greater than 14.
D) The null can be rejected because the majority of the sample values exceed 14.

Answer 1

The null hypothesis cannot be rejected since the test statistic is approximately t = .20,which is not in the rejection region.

Explanation:

Data: 14.25, 13.7, 14.02, 14.13, 13.99, 14.0

Mean =
`(Sum)/(n)=14.015`

Standard deviation =
`\sqrt{\frac{\sum(x-\bar{x})^2}{n}}=0.1836`

Claim : Average number of fluid ounces in the cup should be 14.

Null hypothesis :
`H_0:\mu = 14`

Alternate hypothesis :
`H_a:\mu \\eq 14`

n = 6

`t = (x-\mu)/((s)/(√(n)))\\t=(14.015-14)/((0.1836)/(√(6)))\\t=0.20\\\alpha = 0.05t_(df,\alpha)=t_((5,0.05))=2.571`

t critical > t calculated

So, we failed to reject null hypothesis

So, Option B is true

Hence The null hypothesis cannot be rejected since the test statistic is approximately t = .20,which is not in the rejection region.