In this assignment, I will perform a two tailed test and a one tailed hypothesis test for the average difference in yields that are provided by using old and new fertilizers. The data on yields on 15 parcels of land is provided in the table below:
The first part of the assignment is to perform a two tailed hypothesis test for the two independent samples given. Assume that the population of yields is normally distributed, there are no outliers in the data, the variances are homogeneous. Then, it is possible to use two-sample Student’s t-test for independent samples ("Independent t-test in SPSS Statistics - Procedure, output and interpretation of the output using a relevant example | Laerd Statistics", 2016). Perform the Student’s hypothesis test showing all six steps.
Step 1. State the null and the alternative hypotheses:
Null hypothesis: There is no significant difference between the average yields provided by new fertilizer and old fertilizer.
Alternative hypothesis: There a no significant difference between the average yields provided by new fertilizer and old fertilizer.
Mathematically, these hypotheses can be expressed as follows:
H0: μ1=μ2Ha: μ1≠μ2
Step 2. Select the level of significance. The probability of making a type 1 error is set at the level of 5%.
Step 3. Calculate the critical values. The critical t-value is t(n1+n2-2, 0.05) = t(28, 0.05). According to the Student’s t-table, this value is equal to 2.048.
Step 4. Calculate the statistic. The t-statistics can be calculated as follows ("Psychological Statistics", 2016):
t=x1-x2n1-1s12+n2-1s22n1+n2-21n1+1n2t=14.32-14.0414*0.301714+14*0.27828615+15-2115+115≈1.424
I also used Excel Data Analysis for performing t-test. The results are shown in the table below:
The t-statistic is equal to 1.423933.
Step 5. Compare t-statistic to the t-critical. The t-statistic value is less than the t-critical value.
Step 6. Conclusion and decision. Since the t-statistics value is less than t-critical value, I failed to reject the null hypothesis. There is no enough evidence to say about a significant difference in the average yields provided by the two types of fertilizers (at the 5% level of significance).
The next part of the assignment is to perform a one tailed hypothesis test showing all six steps.
Step 1. State the null and the alternative hypotheses:
Null hypothesis: There is no significant difference between the average yields provided by new fertilizer and old fertilizer.
Alternative hypothesis: The average yield provided by the new fertilizer is significantly higher than the average yield provided by the old fertilizer.
Mathematically, these hypotheses can be expressed as follows:
H0: μ1=μ2Ha: μ1>μ2
Step 2. Select the level of significance. The probability of making a type 1 error is set at the level of 5%.
Step 3. Calculate the critical values. The critical t-value is t(n1+n2-2, 0.05) = t(28, 0.05). Mind that this is a one-tailed value. According to the Student’s t-table, this value is equal to 1.701.
Step 4. Calculate the statistic. The t-statistic is the same, it is equal to 1.423933.
Step 5. Compare t-statistic to the t-critical. The t-statistic value is still less than the t-critical value.
Step 6. Conclusion and decision. Since the t-statistics value is less than t-critical value, I failed to reject the null hypothesis. There is no enough evidence to say that the average yield provided by the new fertilizer is significantly higher than the average yield provided by the old fertilizer (at the 5% level of significance).
Summing up, I should not pay for the new fertilizer if it is more expensive, as there it will give me approximately similar yield. The two-tailed test is stronger, as it gives 5% type 1 error for both directions of difference (for significantly lesser and significantly higher values). The one-tailed test has 5% type one error only for one direction (in our case, for significantly higher values). Consequently, the t-critical of the two-tailed test is higher than of the one-tailed test.
References
Independent t-test in SPSS Statistics - Procedure, output and interpretation of the output using a relevant example | Laerd Statistics. (2016). Statistics.laerd.com. Retrieved 8 September 2016, from https://statistics.laerd.com/spss-tutorials/independent-t-test-using-spss-statistics.php
Psychological Statistics. (2016). Retrieved 8 September 2016, from https://www4.uwsp.edu/psych/stat/11/hyptest2s.htm
