As a part of a study, a chemical plant sampled the amount of chemicals in the soil surrounding their building. Samples were taken in front of the building (sample 1) and samples were taken in the back of the building (sample 2). The results in ppm are listed below. Determine if there is a difference in the amount of chemicals in the soil in the front vs the back of the chemical plant. Perform the five steps of the hypothesis test (assume equal variances).
Sample 1 | Sample 2 |
0 | 5.5 |
1.8 | 3.9 |
1.5 | 5.9 |
0.4 | 2.6 |
3.5 | 6.8 |
1.7 | 1.8 |
3.9 | 4.5 |
5.4 | 2.5 |
2.5 | 3.9 |
2.7 | 3.9 |
Null Hypothesis
Alternative Hypothesis
The value of mean and standard deviation for sample 1 and sample 2 is
Sample 1 | Sample 2 | |
Mean | ||
Standard deviation |
Under H0, the test statistic is
Degrees of freedom = n1+n2-2 = 18
Significance level
The critical value of t for 18 df at 5% suignificance level is +/-2.101
The P-Value is .0232
Since p value is less than significance level. Reject H0.
Hence , at 5% significance level, we have enough evidence to conclude that there is a difference in the amount of chemicals in the soil in the front vs the back of the chemical plant
Get Answers For Free
Most questions answered within 1 hours.