# Population Means Worksheet

 parcel New fertilizer (x) Old fertilizer (y) 1 14.2 -0.02 0.0004 14.0 0.06 0.0036 2 14.1 -0.12 0.0144 13.9 -0.04 0.0016 3 14.5 0.28 0.0784 14.4 0.46 0.2116 4 15.0 0.78 0.6084 14.8 0.86 0.7396 5 13.9 -0.32 0.1024 13.6 -0.34 0.1156 6 14.5 0.28 0.0784 14.1 0.16 0.0256 7 14.7 0.48 0.2304 14.0 0.06 0.0036 8 13.7 -0.52 0.2704 13.7 -0.24 0.0576 9 14.0 -0.22 0.0484 13.3 -0.64 0.4096 10 13.8 -0.42 0.1764 13.7 -0.24 0.0576 11 14.2 -0.02 0.0004 14.1 0.16 0.0256 12 15.4 1.18 1.3924 14.9 0.96 0.9216 13 13.2 -1.02 1.0404 12.8 -1.14 1.2996 14 13.8 -0.42 0.1764 13.8 -0.14 0.0196 15 14.3 0.08 0.0064 14.0 0.06 0.0036 213.3 4.224 209.1 3.896

The new fertilizer tomato mean/average yields is equal to

The old fertilizer tomato mean/average yields is equal to

The new fertilizer yields standard deviation sx =

The old fertilizer yields standard deviation sy =

The two tailed test

H0 : mean of new fertilizer tomato yields = mean of old fertilizer tomato yields

H1 : mean of new fertilizer tomato yields  mean of old fertilizer tomato yields

From the two samples, ,x,sx and ,y, sy are the sample sizes means and standard deviations respectively, then the common or pooled estimates of the standard deviations for both populations, that is Sp can be found as follows.

Sp =   (Terry, 2002)

Sp =  = 0.7337

Sp is then the best estimate of the standard deviation in each population. The standard error of each sample mean can be calculated as follows.

Sx (mean) =  = 0.1894

Sy (mean) =  = 0.1894

And the sampling error of the distribution of differences in sample means is equal to the square root of the sum of the squares of the standard errors of each sample mean, which is equal to 0.2679. The value of the t-score by calculation is given by the difference between the calculated means of the samples divided the sampling error is 1.045 with  degrees of freedom.

The reason why the frugal farmer decided that there was no difference between the two fertilizers is because at 5% level of confidence from the t-distribution table gives 2.048. Since the calculated t-score is less than the t-score from the table, then we accept the null hypothesis and reject the alternative hypothesis.

1. B) The test that should have been done is a one tailed test.

H0 : mean of new fertilizer tomato yields = mean of old fertilizer tomato yields

H1 : mean of new fertilizer tomato yields  mean of old fertilizer tomato yields

This is a one tailed test as the fertilizer sales representative is only interested in the right hand tail of the distribution as calculated in section a above, the t-score at 5% confidence level is 2.048 for a two tailed test and for a one tailed test the t-score is read to be 1.701. As the calculated t-score is less than the 5% value, we accept the null hypothesis. Therefore, as the fertilizer sale representative, it is acceptable for the farmer to conclude that there is no significant difference between the new fertilizer and the old fertilizer that he has been using since the one tailed test does not help to draw a different conclusion.

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1. C) The one tailed test that the sales representative has chosen to use is a stronger and a more reliable test. The reason is that the alternative hypothesis is only concerned with one of the tails of the distribution which is to justify the superiority of the new fertilizer.

Reference

Terry, L. (2002) Quantitative Techniques. London: Cengage Learning.

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