The data set (concrete.xlsx) contains the compressive strength, in thousands of

Question

The data set (concrete.xlsx) contains the compressive strength, in thousands of pounds per square inch (psi), of 30 samples of concrete taken two and seven datys after pouring.

(a) At the 0.05 level of significance, is there evidence of a difference in the mean strength at two days and at seven days?

(b) Find the p-value in (a) ans interpret its meaning.

(c) At the 0.05 level of significance, is there evidence that the mean strength is lower at two days than at seven days?

(d) Find the p-value in (c) and interpret its meaning.

Sample Two Days Seven Days 1 2.830 3.505 2 3.295 3.430 3 2.710 3.670 4 2.855 3.355 5 2.980 3.985 6 3.065 3.630 7 3.765 4.570 8 3.265 3.700 9 3.170 3.660 10 2.895 3.250 11 2.630 2.850 12 2.830 3.340 13 2.935 3.630 14 3.115 3.675 15 2.985 3.475 16 3.135 3.605 17 2.750 3.250 18 3.205 3.540 19 3.000 4.005 20 3.035 3.595 21 1.635 2.275 22 2.270 3.910 23 2.895 2.915 24 2.845 4.530 25 2.205 2.280 26 3.590 3.915 27 3.080 3.140 28 3.335 3.580 29 3.800 4.070 30 2.680 3.805

Explanation / Answer

a)
Hypothesis:
Null hypothesis: mu1 = mu2
Alternative hypothesis: mu1 not equal to mu2

Test statistic;

x1 = 2.9595 , x2 = 3.538 , s1 = 0.4336 , s2 = 0.5139 , n1 = n2 = 30

t = ( x1 - x2) / sqrt(s1^2 / n1 + s2^2/n2)
= ( 2.9595 - 3.538) / sqrt ( 0.4336^2/30 + 0.5139^2/30)
= -4.7124

b)

P value is calculated using t = -4.7124 , df = 58
for two tailed test

p value = .000016.
As p value is less than 0.05 significance level so we reject the null hypothesis.

c)

Hypothesis:
Null hypothesis: mu1 = mu2
Alternative hypothesis: mu1 < mu2

Test statistic;

x1 = 2.9595 , x2 = 3.538 , s1 = 0.4336 , s2 = 0.5139 , n1 = n2 = 30

t = ( x1 - x2) / sqrt(s1^2 / n1 + s2^2/n2)
= ( 2.9595 - 3.538) / sqrt ( 0.4336^2/30 + 0.5139^2/30)
= -4.7124

d)

P value is calculated using t = -4.7124 , df = 58
for one tailed test

p value = .00001
As p value is less than 0.05 significance level so we reject the null hypothesis.

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