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7. The manufacturer of an Engine Energizer System (EES) claims that it improves gas mileage and reduces emissions in automobi
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Answer #1

(a)

The manufacturer of an Engine Energizer System (EES) claims that it proves gas mileage and reduces emissions in automobiles by using magnetic-free energy to increase the amount of oxygen in the fuel for greater combustion efficiency. The test result performed on a random sample of 14 vehicles.

The data of CO (parts per million) levels both before and after installation of EES is given below:

Before After Difference
1.60 0.15 1.45
0.30 0.20 0.10
3.80 2.80 1.00
6.20 3.60 2.60
3.60 1.00 2.60
1.50 0.50 1.00
2.00 1.60 0.40
2.60 1.60 1.00
0.15 0.06 0.09
0.06 0.16 -0.10
0.60 0.35 0.25
0.03 0.01 0.02
0.10 0.00 0.10
0.19 0.00 0.19

The mean CO emission level before installation of EES is,

mean = (\frac{1.60+...+0.19}{14}) = 1.624

The mean CO emission level after installation of EES is,

mean = (\frac{0.15+...+0.00}{14}) = 0.859

Here we want to test whether the average CO emission level get reduced after EES installation.

Hence our null hypothesis of chief interest are:

он : There is no difference between CO emission levels before and after installation of EES. i.e. \mu _{d} = 0

H_{1} : There is significant reduction between CO emission levels before and after installation of EES. i.e. \mu _{d} > 0

Test statistic: t = \frac{\bar{d} - \mu _{d}}{\frac{sd}{\sqrt{n}}}

Here \bar{d} = The sample mean difference = 1.624 - 0.859 = 0.765

\mu _{d} = Poplulation mean difference = 0 (under он )

sd = Standard deviation of the differences = 0.909

\frac{sd}{\sqrt{n}} = 0.909/\sqrt{14} = 0.243

Then  t = \frac{0.765 - 0}{0.243} = 3.148

The critical value of t is t_{0.01,13} = 2.650

Here t_{obs} =3.148

So, t_{obs} > 2.65

Ans: Hence, we can conclude that there is significant reduction in CO emission levels after installation of EES.

(b)

The confidence interval around the difference in means

\bar{d} \pm (t_{\alpha , n-1} * \frac{sd}{\sqrt{n}})

Here,

\bar{d} = The sample mean difference = 1.624 - 0.859 = 0.765

\mu _{d} = Poplulation mean difference = 0 (under он )

sd = Standard deviation of the differences = 0.909

\frac{sd}{\sqrt{n}} = 0.909/\sqrt{14} = 0.243

The critical value of t is t_{0.01,13} = 2.650

Hence, \bar{d} \pm (t_{\alpha , n-1} * \frac{sd}{\sqrt{n}}) = 1.624 \pm (2.650*0.243) = (0.98, 2.268)

The confidence interval around the difference in means is (0.98, 2.268)

(c)

Ans: there is a 99% chance that the confidence interval (0.98, 2.268) we calculated contains the true population mean difference i.e. if we take samples 100 times from this population then 99% of the times the confidence interval (0.98, 2.268) we calculated contains the true population mean difference.

Since the sample mean difference = 0.765 does not lie in the confidence interval of (0.98, 2.268) we reject the null hypothesis that the true population mean difference is 0 and conclude that the true population mean difference is not 0 i.e. there is a significant reduction in CO emission levels after installation of EES.

Ans: The interpretation of confidence interval supports the conclusion of hypothesis test in part (a).

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