An article reports the following data on burn time (hours) for samples of oak an

ID: 3225695 • Letter: A

Question

An article reports the following data on burn time (hours) for samples of oak and pine. Test at level 0.05 to see whether there is any difference in true average burn time for the two types of wood.

State the appropriate null and alternative hypotheses. (Let 1 and 2 denote the true average burn times for pine and oak, respectively.)

H0: 1 2 = 0
Ha: 1 2 < 0H0: 1 2 = 0
Ha: 1 2 > 0 H0: 1 2 = 0
Ha: 1 2 0H0: 1 2 = 0
Ha: 1 2 0

Calculate the test statistic.
w =

State the conclusion in the problem context.

Fail to reject H0. There is no evidence of a difference in true average burn time for the two types of wood.

Fail to reject H0. There is evidence of a difference in true average burn time for the two types of wood.

Reject H0. There is evidence of a difference in true average burn time for the two types of wood.

Reject H0. There is no evidence of a difference in true average burn time for the two types of wood.

Oak 1.70 0.62 1.57 1.61 1.39 1.24 1.82 0.52 Pine 0.93 1.35 1.31 1.55 0.73 1.20

Explanation / Answer

Let,

X1 = Oak

X2 = Pine

1.7

0.93

0.62

1.35

1.57

1.31

1.61

1.55

1.39

0.73

1.24

1.2

1.82

0.52

Mean =

1.3088

1.1783

Standard deviation =

0.4901

0.2994

Null and Alternative Hypothesis:

Ha: µ1 - µ2 ¹ 0 H0: µ1 - µ2 = 0

Test Statistics:

t = (x1 bar – x2 bar )/SE (x1-x2)

SE (x1-x2) = ((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2) * (1/n1+1/n2)

n1 = 8

X1 bar = 1.3088

s1 = 0.4901

n2 = 6

X2 bar = 1.1783

s2 = 0.2994

SE (x1-x2) = (8-1)*0.4901^2+(6-1)*0.2994^2)/(8+6-2) * ((1/8)+(1/6))

= 0.2275

t = (1.3088 – 1.1783 )/0.2275

= 0.574

W = 0.574

Now for 12 degrees of freedom at 5% level of significance from t-table we get the critical t as -2.179 and + 2.179. The obtained test statistics falls in this interval of -2.179 and + 2.179.

We fail to reject the null hypothesis.