Question

1. two types of fertilizer should be compared. For this purpose, 25 plots of the same size are fertilised, namely N=10 plots of Type A and M=15 plots of Type B. for the former, The average yield is 23.6 with sample variance 9.5 and for the other plots, the average is 20.1 with variance 8.9. check a) homogeneity and B) the equality of the mean values with a significance level of 10% either with parametric test methods or also with nonparametric methods.

Answer 1

(A) test for homogenity:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the the rejection region for this two-tailed test is R={F: F>2.646}.

3.test statistic

The provided sample variances are ​and and the sample sizes are given by n1​=10 and n2=15.

The F-statistic is computed as follows:

(4) Decision about the null hypothesis

Since from the sample information we get that F=1.067<F=2.646, it is then concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance ​ is different than the population variance at the α=0.1 significance level.

(B)test of equality of means:

first we find the standard deviations:

s1=

s2=

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the degrees of freedom are df=23. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc=1.714, for α=0.1 and df=23

The rejection region for this two-tailed test is R={t:∣t∣>1.714}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t=

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=2.837>tc​=1.714, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0093, and since p=0.0093<0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 is different than μ2​, at the 0.1 significance level.

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