Question

The Local Bottling Company has just installed a new bottling process that will fill 16-ounce bottles of the popular Paradise Classic Cola soft drink. Both overfilling and underfilling bottles are undesirable: Underfilling leads to customer complaints and overfilling costs the company considerable money. In order to verify that the filler is set up correctly, the company wishes to see whether the mean bottle fill, μ, is close to the target fill of 16 ounces. To this end, a random sample of 38 filled bottles is selected from the output of a test filler run. If the sample results cast a substantial amount of doubt on the hypothesis that the mean bottle fill is the desired 16 ounces, then the filler’s initial setup will be readjusted.

(a) The bottling company wants to set up a hypothesis test so that the filler will be readjusted if the null hypothesis is rejected. Set up the null and alternative hypotheses for this hypothesis test.

H₀ : μ

H_a : μ

(b) Suppose that Local Bottling Company decides to use a level of significance of α = 0.01, and suppose a random sample of 38 bottle fills is obtained from a test run of the filler. For each of the following two sample means— x⎯⎯

= 16.05, x⎯⎯

= 15.97} — determine whether the filler’s initial setup should be readjusted. In each case, use a critical value, a p-value, and a confidence interval. Assume that σ equals .1. (Round your z to 2 decimal places and p-value to 4 decimal places.)

x⎯⎯

= 16.05

z
p-value

Decision:

x⎯⎯

= 15.97

z
p-value

DECISION:

NEED ANSWER IN NEXT 20 MIN TIMED HW

Answer 1

(a)

H0: Null Hypothesis: = 16

HA: Alternative Hypothesis: 16

(b)

(i)

n = 38

= 16.05

= 0.1

SE = /

= 0.1/

= 0.0162

Z = (16.05 - 16)/0.0162

= 3.0822

Table of area Under Standard Normal Curve gives area = 0.4990

So,

P - Value = (0.5 - 0.4990) X 2 = 0.0020

Since P - Value is less than , the difference is significant. Reject null hypothesis.

Conclusion:

The filler's initial set up should be readjusted.

So,

Answers are:

Z	3.08
P - Value	0.0020
Decision	The filler's initial set up should be readjusted

(ii)

n = 38

= 15.97

= 0.1

SE = /

= 0.1/

= 0.0162

Z = (15.97 - 16)/0.0162

= - 1.85

Table of area Under Standard Normal Curve gives area = 0.4678

So,

P - Value = (0.5 - 0.4678) X 2 = 0.0644

Since P - Value is greater than , the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The filler's initial set up should be readjusted.

So,

Answers are:

Z	- 1.85
P - Value	0.0644
Decision	The filler's initial set up need not be readjusted