Question

A drink bottler has several devices that pour a specific amount of liquid into each bottle. They test whether the temperature of the liquid has an impact on how much the devices pour. From a sample of the devices, each is tested twice: once with cold liquid, and once with warm. Is there evidence of a difference in the amounts poured at different temperatures? a) What kind of test is this? What are the hypotheses? c) What conditions must be satisfied? d) Assume the sample is representative. Find the p-value and give your conclusion in context.

Warm
21.2
20
19.7
19.7
20.3
20
20.6
17.6
18.4
19.6

Cold
20.6
20.5
19.5
20.1
20.7
19.8
19.9
18.4
19.2
21.3

Answer 1

From the data, the following was found

	Warm	Hot
n	10	10
Sum	197.1	200
Average	19.71	20
SS(Sum of squares)	9.709	301.1
Variance = SS/n-1	1.079	33.456
Std Dev=Sqrt(Variance)	1.04	5.78

(a) The test used is an independent sample t test

(b) The Hypothesis

H0: = : The mean amounts of liquids poured at hot and cold temperatures are the same.

Ha: : The mean amounts of liquids poured at hot and cold temperatures are different.

Assumptions:

(a) The sample is a simple random sample.

(b) The samples are independent of each other.

(c) The samples come from normal populations or approximately normal populations

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To find the p value, we need to calculated the degrees of freedom (for assuming unequal variances) and the test statistic.

Degrees of Freedom is calculated as

Substituting the values, we get df = 10

The Test Statistic:

t = (Difference in means) / SE

SE = SQRT [(s1)² / n1 + (S2)² / n2] = 1.85715

Therefore t = (19.71 - 20) / 1.85715 = -0.16

The p value (2 tailed) for t = -0.16, df = 10 is 0.8761

Our conclusion is : Do not Reject H0. There is not sufficient evidence to conclude that the mean amounts of liquids poured at hot and cold temperatures are different.

________________________________________________________________________

__________________________________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

Warm	Mean	(X - Mean)^2		Cold	Mean	(X - Mean)^2
21.2	19.71	2.2201		20.6	20	0.36
20	19.71	0.0841		20.5	21	0.25
19.7	19.71	0.0001		19.5	22	6.25
19.7	19.71	0.0001		20.1	23	8.41
20.3	19.71	0.3481		20.7	24	10.89
20	19.71	0.0841		19.8	25	27.04
20.6	19.71	0.7921		19.9	26	37.21
17.6	19.71	4.4521		18.4	27	73.96
18.4	19.71	1.7161		19.2	28	77.44
19.6	19.71	0.0121		21.3	29	59.29
197.1	SS	9.709		200	SS	301.1
19.71	Var	1.078777778		20	Var	33.45555556
	SD	1.038642276			SD	5.784077762