Question

The “cold start ignition time” of an automobile engine is investigated by a gasoline manufacturer. The following 8 times (in seconds) were obtained for a test vehicle:

1.75, 1.92, 2.62, 2.35, 3.09, 3.15, 2.53, 1.91

A Construct and interpret the 95% confidence interval for the population mean start ignition time of an automobile engine.

B Test at 0.05 significance level whether the population mean start ignition time is below 2.9 
seconds. Include the hypotheses, the test st atistic, the p-value, test decision 
and conclusion in the context of the problem.

C. Does the confidence interval in the output support your statistical decision? Explain.

D. Construct a Normal Probability Plot of the data. In your opinion, does the graph support an assumption of normality? Explain.

E. Perform a Shapiro-Wilk test of normality. State hypothesis, p-value and conclusion. Did this support your decision in part (d) based on the normality plot?

F. What test decision error could you have made and provide an explanation of this error in context of the problem.

G. Include a copy of your R-code, test output, and normal probability plot.

Answer 1

(a)

Following table is used to calculate the sample mean and sample standard deviation.

95% confidence interval of the mean:

Degree of freedom: n-1 = 8-1=7

CI = Mean ± (tcrit X s/sqrt(n))

= 2.415 ± (2.365 x 0.534 / sqrt(8))

=(1.969, 2.861)

The 95% confidence interval is (1.969, 2.861)

(b)

H0:μ=2.9

Ha:μ<2.9

It is one-tailed test. Rejection region: If the p-value is less than the significance level, we reject the null hypothesis and accept the alternative hypothesis.

t = (mean - value mean)/(s/sqrt(n))
= (2.415 - 2.9)/(0.534/sqrt(8))
= -2.5689

P (T ≤ t) = P (T ≤ -2.5689)
= 0.0185

Hence, there is evidence to conclude that the population mean start ignition time is below 2.9 seconds.

(c)  The confidence interval is (1.969,2.861). Since the null value is not in this interval, we conclude that we are 95% confident that the population mean start ignition time is definitely below 2.9 seconds. This also matches with the test at 0.05 significance level that the population mean start ignition time is below 2.9 seconds.

(d)

Computing for (Sample no-0.375)/(Total samples + 0.25) to get probability

and NORMSINV(probability), we get the graph below based on the data

1.75	1	0.075758	-1.4342
1.92	2	0.19697	-0.8525
2.62	3	0.318182	-0.47279
2.35	4	0.439394	-0.15251
3.09	5	0.560606	0.152506
3.15	6	0.681818	0.472789
2.53	7	0.80303	0.852495
1.91	8	0.924242	1.4342

(e)

use the code

x<-c(1.75,1.91,1.92,2.35,2.53,2.62,3.09,3.15)
shapiro.test(x)
hist(x,main="Main",xlab="value",border="light blue",col="blue",las=1)

to get the results

Skewness: 0.2128

P-value: 0.4382

(f) the error here is that we are not considering the null and alternate hypothesis as we considered in previous part.