Question

QUESTION A new ferrorcom has advertised that can significantly increase crepe A group of researchers w h o testif this is true. They random assign the few to parts of their field. Once the harvest has come they determine that the mean crop yield for the part of helds that had the forte is 65 with a standardevono 100. The 32 parts of the fields that did not receive the ferter had a man rod of 601 with a standard deviation of Does it appear that the new fertiler significantly increases croped 0.01) • Hoppu HA DI>DN HUN HAN HA HAN-UN • He Ho UN HAWW QUESTION 10 What the stand For the toolbar, presSALTI TTTA for the previous problemCearylableach ALTINF10 M T.

Answer 1

Solution-

Let,

X1 = Part of the field that did not receive fertilizer

X2 = Part of the field that had fertiliser.

● last option has correct hypothesis

Ho: μf = μN

HA: μ​f  < μN

◆ Test Results-

The provided sample means are shown below: X1 = 601 X2 = 675 Also, the provided sample standard deviations are: $1 = 90 S2 = 100 and the sample sizes are nı = 32 and n2 = 36. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = M2 Ha: M1 < M2 This corresponds to a left-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 a = 0.01, and the degrees of freedom are df = 66. 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 left-tailed test is te = –2.384, for a = 0.01 and df = 66. The rejection region for this left-tailed test is R = {t:t < -2.384}.

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - 72 „(m11) s{+(12-1) si ( 1 + 1) ni+n2-2 = -3.192 601 - 675 (32-1)902+(36-1)1002 - 32+36- 2 3 2 + 36) Plo (4) Decision about the null hypothesis Since it is observed that t = -3.192 <te= -2.384, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0011, and since p= 0.0011< 0.01, 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 mi is less than M2, at the 0.01 significance level. Confidence Interval The 99% confidence interval is –135.499 < MI – M2 < -12.501. Graphically. T-Test: t= -3.192, p = 0.0011 0.40 0.35 0.30 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Test Results that the null hypothesis is rejected.

So there is enough evidence to claim that the new fertilizer significantly increase the crop yield.