The breaking strength of hockey stick shafts made of two different graphite-Kevl

Question

The breaking strength of hockey stick shafts made of two different graphite-Kevlar composites yield the following results (in newtons):

Composite A: 487.3   444.5   467.7   456.3   449.7   459.2   478.9   461.5 477.2

Composite B: 488.5   501.2   475.3   467.2   462.5   499.7   470.0   469.5 481.5

                          485.2   509.3   479.3   478.3   491.5

a) Assuming normality, can you conclude that the mean breaking strength is smaller for hockey sticks made from composite B by at least 2 newtons? Carry out the appropriate test by hand.

b) If the investigator new of similar population variances, use the data given in Exercise 4 to test that the mean breaking strength differs between the 2 composites. Do this by hand. Then verify your work with the appropriate confidence interval.

Explanation / Answer

Part a

Here, we have to use two sample t test for the population means assuming unequal population variances. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: Mean breaking strength for composite A is smaller for hockey sticks made from composite B by at least 2 Newton.

Alternative hypothesis: Ha: Mean breaking strength for composite A is smaller for hockey sticks made from composite B by less than 2 Newton.

H0: µB - µA 2 versus Ha: µB - µA < 2

This is a lower tailed test.

Test statistic formula is given as below:

t = ((X1bar – X2bar) - µd) / sqrt[(S1^2/n1)+(S2^2/n2)]

From the given data, we have

X1bar = 483.1307692

X2bar = 464.7

S1 = 14.4491

S2 = 14.2379

n1 = 13

n2 = 9

S1^2 = 208.7756

S2^2 = 202.7175

S1^2/n1 = 16.0597

S2^2/n2 = 22.5242

Numerator df = ((S1^2/n1)+( S2^2/n2))^2 = (16.0597 + 22.5242)^2 = 1488.717339

Denominator df = ((S1^2/n1)^2/(n1 – 1)) + ((S2^2/n2)^2/(n2 – 1))

Denominator df = 16.0597^2/(13 – 1) + 22.5242^2/(9 – 1) = 84.9100

Total df = numerator df / denominator df = 1488.7120/ 84.9100 = 17.5328

df = 17

Standard error = sqrt[(S1^2/n1)+(S2^2/n2)]

Standard error = sqrt(16.0597 + 22.5242)

Standard error = 6.2116

(X1bar – X2bar) = 483.1307692 - 464.7 = 18.4308

t = ((X1bar – X2bar) - µd) / sqrt[(S1^2/n1)+(S2^2/n2)]

µd = 2

t = (18.4308 – 2) / 6.2116

t = 2.6452

P-value = 0.9915

= 0.05

P-value >

So, we do not reject the null hypothesis that Mean breaking strength for composite A is smaller for hockey sticks made from composite B by at least 2 Newton.

There is sufficient evidence to conclude that Mean breaking strength for composite A is smaller for hockey sticks made from composite B by at least 2 Newton.

Part b

We have to conduct same test by using equal variances t test.

H0: µB - µA 2 versus Ha: µB - µA < 2

This is a lower tailed test.

Test statistic formula is given as below:

t = ((X1bar – X2bar) - µd) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From the given data, we have

X1bar = 483.1307692

X2bar = 464.7

S1 = 14.4491

S2 = 14.2379

n1 = 13

n2 = 9

S1^2 = 208.7756

S2^2 = 202.7175

DF = n1 + n2 – 2 = 13+9 – 2 = 22 – 2 = 20

Sp2 = [(13 – 1)* 208.7756 + (9 – 1)* 202.7175]/(13 + 9 – 2)

Sp2 = 206.3524

t = ((X1bar – X2bar) - µd) / sqrt[Sp2*((1/n1)+(1/n2))]

t = (18.4308 – 2) /sqrt[206.3524*((1/13)+(1/9))]

t = (18.4308 – 2) / 6.2291

t = 2.6378

P-value = 0.9921

= 0.05

P-value >

So, we do not reject the null hypothesis that Mean breaking strength for composite A is smaller for hockey sticks made from composite B by at least 2 Newton.

There is sufficient evidence to conclude that Mean breaking strength for composite A is smaller for hockey sticks made from composite B by at least 2 Newton.

95% confidence interval is given as below:

Degrees of freedom = DF = n1 + n2 – 2 = 13+9 – 2 = 22 – 2 = 20

Critical t value = 2.0860

Standard error = sqrt[Sp2*((1/n1)+(1/n2))]

Standard error = 6.2291

Margin of error = critical value * standard error

Margin of error = 2.0860*6.2291 = 12.9936

(X1bar – X2bar) = 483.1307692 - 464.7 = 18.4308

Confidence interval = (X1bar – X2bar) -/+ margin of error

Confidence interval = 18.4308 -/+ 12.9936

Lower limit = 18.4308 - 12.9936 = 5.4372

Upper limit = 18.4308 + 12.9936 = 31.4244

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