Given: In order to evaluate the suitability of nonpotable water available at the
ID: 3273912 • Letter: G
Question
Given: In order to evaluate the suitability of nonpotable water available at the job site for mixing concrete, six standard mortar cubes were made using the water and six others using potable water. The cubes were tested for compressive strength after 7 days of curing and produced the loads to failure (in pounds) shown in the the table. Based on these results only, would you accept that water for mixing concrete? (assume the level of significance is 0.05).
Cubes Made with Non-potable Water
Cubes Made with Potable Water
14530
16530
15820
16820
14160
17920
16100
15690
15870
18820
14380
16720
Required: Null and alternative hypothesis stated. Tcalc. Tvalue. T-test claim correct.
Cubes Made with Non-potable Water
Cubes Made with Potable Water
14530
16530
15820
16820
14160
17920
16100
15690
15870
18820
14380
16720
Explanation / Answer
Null Hypothesis H0: Mean loads to failure for cubes made with non-potable Water is equal to mean loads to failure for cubes made with potable Water.
Alternative Hypothesis H1: Mean loads to failure for cubes made with non-potable Water is less than the mean loads to failure for cubes made with potable Water.
Loads to failure for cubes Made with Non-potable Water = (14530, 15820, 15820, 15820, 15820, 15820)
Loads to failure for cubes Made with potable Water = (16530, 16530, 17920, 15690, 18820, 16720)
Mean of loads to failure for cubes Made with Non-potable Water = 15605
Mean of loads to failure for cubes Made with potable Water = 17035
Standard deviation of loads to failure for cubes Made with Non-potable Water = 526.64
Standard deviation of loads to failure for cubes Made with Potable Water = 1130.30
The standard error (SE) of the sampling distribution.
SE = sqrt[ (s12/n1) + (s22/n2) ]
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
SE = sqrt[ (526.642/6) + (1130.302/6) ] = 509.07
The degrees of freedom (DF) is:
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (526.642/6 + 1130.302/6)2 / { [ (526.642 / 6)2 / (6 - 1) ] + [ (1130.302 / 6)2 / (6 - 1) ] }
= 7 (Rounding to nearest integer)
T statistic = (Mean of sample 1 - Mean of sample 2) / SE
= (526.64 - 1130.30) / 509.07 = - 1.186
P-value for T = - 1.186 at DF = 7 is 0.1372
As, the p-value is greater than level of significance is 0.05, we fail to reject the null hypothesis and conclude that the mean loads to failure for cubes made with non-potable Water is equal to mean loads to failure for cubes made with potable Water. So, we can claim that the nonpotable water available at the job site are suitable for mixing concrete.
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