Question

A substance used in biological and medical research is shipped by airfreight to users in cartons of 1,000 ampules. The data in the Excel file Quiz3.xlsx (Tab Air), involving 10 shipments, were collected on the number of times the carton was transferred from one aircraft to another over the shipment route (x) and the number of ampules found to be broken upon arrival (y). Assume that the first-order regression model given below is appropriate:

y = beta0 + beta1 x + e

y	x
16	1
9	0
17	2
12	0
22	3
13	1
8	0
15	1
19	2
11	0

Conduct a t-test to decide whether or not there is a linear association between number of times a carton is transferred (x) and number of broken ampules (y). Use a level of significance of 0.05. Clearly write every step of the hypothesis testing procedure.
Estimate the mean breakage for 4 transfers by constructing a 99% confidence interval. Interpret the confidence interval. Round your answer to 2 decimal places.
The next shipment will entail two transfers. Obtain a 99% prediction interval for the number of broken ampules for this shipment. Interpret this confidence interval. Round your answer to 2 decimal places.

Answer 1

Regression y on x, we would obtain a causal linear relationship between the two variables, expressed in the form:

y hat = a + bx

=Predicted y

where the intercept coefficient is estimated by the formula:

a = y bar - b(x bar) and slope coefficient estimate can be obtained using the formula:

Computing the values,

Y	X	XY	X2
16	1	16	1
9	0	0	0
17	2	34	4
12	0	0	0
22	3	66	9
13	1	13	1
8	0	0	0
15	1	15	1
19	2	38	4
11	0	0	0
Y bar = 14.2	X bar = 1	Sum (XY) = 182	Sum (x2) = 20
Sum (Y) = 142	Sum (X) = 10

we get Slope estimate = [(10)(182) - (142)(10) ] / [10(20) - (10)2]

= 4

And intercept estimate = 14.2 - (4)(1)

= 10.2

Hence, the fitted regression equation is expressed as:

y hat = 10.2 + 4x

Stating the hypothesis:

To test whether or not there is a linear association between number of times a carton is transferred (x) and number of broken ampules (y),

H0: b = 0 Vs Ha: b is not equal to zero.

Computing the test statistic:

Using one sample t test,

t = b / SE(b) ~ tn-2

with critical region |t| >tn-2

where SE (b) = Standard error of slope can be computed using the formula:

SE (b) = sqrt [ Σ(yi – ŷi)2 / (n – 2) ] / sqrt [ Σ(xi – x)2 ]

Y	X	y hat	(y-y hat)2
16	1	14.2	3.24
9	0	10.2	1.44
17	2	18.2	1.44
12	0	10.2	3.24
22	3	22.2	0.04
13	1	14.2	1.44
8	0	10.2	4.84
15	1	14.2	0.64
19	2	18.2	0.64
11	0	10.2	0.64
Y bar = 14.2	X bar = 1	Sum [(y - y hat)^2]	17.6
Sum (Y) = 142	Sum (X) = 10

SE (b) = sqrt [ 17.6 / (10 – 2) ] / sqrt [ 10 ]

= 0.469

Substituting in t:

t = 4 / 0.469 = 8.528

Comparing the test statistic obtained with the critical value:

Critical t value for 10 - 2 = 8 degrees of freedom at alpha = 0.05, from t table:

tCritical = 2.306

Decision on whether or not to reject the null:

Since, t = 8.528 > 2.306 lies in the critical / rejection refio, we may reject H0 at 5% level.

Conclusion:

We may conclude that the there is a linear association between number of times a carton is transferred (x) and number of broken ampules (y)

For x = 4,

Predicted y = 10.2 + 4(4)

= 26.2

99% confidence interval for mean breakage for 4 transfers can be obtained as follows:

Predicted y minus / plus E,

where

Substituting the values,

Hence, the interval is obtained as:

(26.2 - 4.9768, 26.2 + 4.9768)

= (21.22, 31.18)

We may interpret this interval as: For a shipment that entails four transfers, the number of broken ampules would lie in the range (21.22, 31.18) with 99% confidence.

Next to compute the Prediction interval for x = 2:

Y hat = 10.2+4(2) = 18.2

y hat plus / minus E

where,

Substituting the values,

Hence, the interval is obtained as:

(18.2 - 5.4519, 18.2 + 5.4519)

= (12.7481,23.6519)

We may interpret this interval as: If the next shipment will entail two transfers, the number of broken ampules is predicted to lie in the range (12.75,23.65) with 99% confidence.