An article contains measurements on samples of coal from sev

An article contains measurements on samples of coal from several counties from Kentucky. In units of percent ash, five samples from Knott County had an average aluminum dioxide (AlO₂) content of 32.17 and a standard deviation of 2.23. Six samples from Lesile County had an average AlO₂ content of 26.48 and a standard deviation 2.02.

A. Can we conclude that the variance of AlO₂ content of coal sample differs between two countries?

i. State hypothesis.

ii. Compute P-value

iii. What is your conclusion?

B. Can you conclude that the mean of AlO₂ contend of coal sample differs between two counties?

i. State hypothesis.

ii. Compute P-value

iii. What is your conclusion?

C. Find a 95% confidence interval for the difference in AlO₂ contend of coal samples from the two counties.

Solution

A. Can we conclude that the variance of AlO₂ content of coal sample differs between two countries?

i. State hypothesis.

Let o1^2 be the variance for Knott County

Let o2^2 be the variance for Lesile County

Null hypothesis: o1^2=o2^2

Alternative hypothesis: o1^2 not equal to o2^2

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ii. Compute P-value

The test statistic is

F=s1^2/s2^2

= 2.23^2/2.02^2

=1.22

The p-value= P(F with df1=n1-1=5-1=4, df2=6-1=5 >1.22) =0.4072 (from F table)

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iii. What is your conclusion?

Assume that the significant level a=0.05

Since the p-value is larger than 0.05, we do not reject the null hypothesis.

SO we can not conclude that the variance of AlO₂ content of coal sample differs between two countries

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B. Can you conclude that the mean of AlO₂ contend of coal sample differs between two counties?

i. State hypothesis.

Let mu1 be the mean for Knott County

Let mu2 be the mean for Lesile County

Null hypothesis: mu1=mu2

Alternative hypothesis: mu1 not equal to mu2

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ii. Compute P-value

The test statisitc is

t=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(32.17-26.48)/sqrt(2.23^2/5+2.02^2/6)

=4.40

The degree of freedom =n1+n2-2=5+6-2=9

It is a two-tailed test.

So the p-value = 2*P(t with df=9 >4.40) =0.0017 (from student t table)

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iii. What is your conclusion?

Since the p-value is less than 0.05, we reject the null hypothesis.

So we can conclude that the mean of AlO₂ contend of coal sample differs between two counties

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C. Find a 95% confidence interval for the difference in AlO₂ contend of coal samples from the two counties.

Given a=1-0.95=0.05, t(0.025, df=9) =2.26 (from student t table)

So the lower bound is

(xbar1-xbar2) -t*sqrt(s1^2/n1+s2^2/n2)

=(32.17-26.48)-2.26*sqrt(2.23^2/5+2.02^2/6)

=2.765376

So the upper bound is

(xbar1-xbar2) +t*sqrt(s1^2/n1+s2^2/n2)

=(32.17-26.48)+2.26*sqrt(2.23^2/5+2.02^2/6)

=8.614624