Question

Urgent

Plot	Fertilizer A	Fertilizer B	Fertilizer C
1	563	588	575
2	593	624	593
3	542	576	564
4	649	672	653
5	565	583	556
6	587	612	590
7	595	617	607
8	429	446	423
9	500	515	483
10	610	641	626
11	524	547	523
12	559	586	568
13	546	582	551
14	503	530	502
15	550	573	567
16	492	518	495
17	497	529	513
18	619	643	626
19	473	497	479
20	533	556	540

f) Calculate a 90% confidence interval estimate of applying fertilizer B data for crop yield in all plots of lands and interpret your results. g) Assuming the population mean of crop yield is 570 bushels for all fertilizers with a standard deviation of 40 bushels for all. Formulate a test hypothesis that crop yield by applying fertilizer C differs from the population crop yield for all fertilizers. Conduct the hypothesis test, conclude your analysis, and explain your answer. Use both critical value and p-value approach with alpha=0.05 h) Calculate a 90% confidence interval estimate of the difference between the population mean yield of fertilizers B and A. Can we conclude at 0.05 level of significance, that the crop yield using fertilizer B is greater than the crop yield using fertilizer A? (hint: you can use the template in chapter 10 to calculate degrees of freedom and the standard error) i) If we assume that observations are now plots of lands, can the scientist infer that there are differences between the three types of fertilizers?

Answer 1

f)

One-Sample T: Fertilizer B

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
20	571.8	55.7	12.5	(545.7, 597.8)

μ: mean of Fertilizer B

The crop yield due to fertilizer B will lie between (545.7, 597.8) with 95% probability and mean value 571.8.

g)

One-Sample Z: Fertilizer C

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
20	551.70	57.79	8.94	(534.17, 569.23)

μ: mean of Fertilizer C
Known standard deviation = 40

Test

Null hypothesis	H₀: μ = 570
Alternative hypothesis	H₁: μ ≠ 570

Z-Value	P-Value
-2.05	0.041

P-value is less than 0.05 so we reject H0. Also we reject H0 if |Z| > 1.96 which is true as |Z| = 2.05 > 1.96. Hence reject H0. We conclude based on the available evidence (data) that crop yield due to fertilizer C is significantly less than the populatuon yield.

Before comparing the means of yield for fertilzer B and A using t test we go for comaring the equality of variances. The output is given below.

Test and CI for Two Variances: Fertilizer B, Fertilizer A

Method

σ₁: standard deviation of Fertilizer B

σ₂: standard deviation of Fertilizer A

Ratio: σ₁/σ₂

The Bonett and Levene's methods are valid for any continuous distribution.

Descriptive Statistics

Variable	N	StDev	Variance	95% CI for σ
Fertilizer B	20	55.715	3104.197	(41.629, 82.670)
Fertilizer A	20	54.350	2953.945	(41.089, 79.702)

Ratio of Standard Deviations

Estimated Ratio	95% CI for Ratio using Bonett	95% CI for Ratio using Levene
1.02512	(0.623, 1.663)	(0.590, 1.707)

Test

Null hypothesis	H₀: σ₁ / σ₂ = 1
Alternative hypothesis	H₁: σ₁ / σ₂ ≠ 1
Significance level	α = 0.05

Method	Test Statistic	DF1	DF2	P-Value
Bonett	0.01	1		0.910
Levene	0.00	1	38	0.963

We see that p-values are too large so we conclude that variances are not significantly different.

Now following is the output of 90% confidence interval for the difference between the average yield of fertilizer B and A.

Estimation for Difference

Difference	Pooled StDev	90% CI for Difference
25.3	55.0	(-4.0, 54.6)

Now following is the result of t test at 0.05 level of significance.

Two-Sample T-Test and CI: Fertilizer B, Fertilizer A

Method

μ₁: mean of Fertilizer B

µ₂: mean of Fertilizer A

Difference: μ₁ - µ₂

Equal variances are assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Fertilizer B	20	571.8	55.7	12
Fertilizer A	20	546.5	54.4	12

Estimation for Difference

Difference	Pooled StDev	95% Lower Bound for Difference
25.3	55.0	-4.0

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ > 0

T-Value	DF	P-Value
1.45	38	0.077

We can observe p value = 0.077 > 0.05, therefore we do not reject H0 at 0.05 level of significance and conclude that there not enough evidence to infer that fertilzer B produces more yield that fertilizer A on average.