Question

2. A plant distills liquid air to produce oxygen, nitrogen, and argon. The percentage of impurity in the oxygen is thought to be linearly related to the amount of impurities in the air as measured by the pollution count" in part per million (ppm). A sample of plant operating data is shown below. Pollution count (ppm) Purity (%) 933 1145 1.59 92.4 91.7 94 1.08 Purity (%) 946 12 93.6 0.99 0.83 12 932 1.47 1.81 05 115 Pollution count (ppm) 90.1 Figure 2. Scatter diagram a) Determine the estimated regression line.

this is concept.

Answer 1

(a) y = 96.4546 - 2.9010*x

(b) 87.4% of the variation in the model is explained.

(c) The hypothesis being tested is:

H0: β1 = 0

H1: β1 ≠ 0

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the slope is significant.

(d) Fitted value = 90.7977

Pollution count	Purity
1.1	93.3
1.45	92
1.36	92.4
1.59	91.7
1.08	94
0.75	94.6
1.2	93.6
0.99	93.1
0.83	93.2
1.22	92.9
1.47	92.2
1.81	91.3
2.03	90.1
1.75	91.6
1.68	91.9
	r²	0.874
	r	-0.935
	Std. Error	0.428
	n	15
	k	1
	Dep. Var.	Purity
ANOVA table
Source	SS	df	MS	F	p-value
Regression	16.4908	1	16.4908	90.13	3.28E-07
Residual	2.3786	13	0.1830
Total	18.8693	14
Regression output				confidence interval
variables	coefficients	std. error	t (df=13)	p-value	95% lower	95% upper
Intercept	96.4546
Pollution count	-2.9010	0.3056	-9.494	3.28E-07	-3.5611	-2.2408
Predicted values for: Purity
		95% Confidence Interval	95% Prediction Interval
Pollution count	Predicted	lower	upper	lower	upper	Leverage
1.95	90.7977	90.3376	91.2578	89.7654	91.8300	0.248

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