Question

Fertilizer: In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Twelve randomly selected plots of land were treated with fertilizer A, and

7

randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results.

Fertilizer A
464	483	441	491	403	466
448	457	437	516	417	420

Fertilizer B
362	414	412	398	382
377	393

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Part 1 of 3

Your Answer is correct

(a) Explain why it is necessary to check whether the populations are approximately normal before constructing a confidence interval.

Since the sample size is ▼small, it is necessary to check that the populations are approximately normal.

Part 2 of 3

Your Answer is correct

(b) Following are boxplots of these data. Is it reasonable to assume that the populations are approximately normal?

400

420

440

460

480

500

520

540

360

370

380

390

400

410

420

It ▼is reasonable to assume that the populations are approximately normal.

Part: 2 / 3

2 of 3 Parts Complete

Part 3 of 3

(c) Construct an

80%

confidence interval for the difference between the mean yields for the two types of fertilizer. Let

μ1

denote the mean yield for fertilizer A. Use tables to find the critical value and round the answer to one decimal place.

The 80% confidence interval for the difference between the mean yields for the two types of fertilizer is <<−μ1μ2 .

Answer 1

Fertilizer A	Fertilizer B
464	362
483	414
441	412
491	398
403	382
466	377
448	393
457
437
516
417
420
Fertilizer A		Fertilizer B
Mean	453.5833333	Mean	391.1428571
Standard Error	9.504750274	Standard Error	7.139523031
Median	452.5	Median	393
Mode	#N/A	Mode	#N/A
Standard Deviation	32.92542078	Standard Deviation	18.88940242
Sample Variance	1084.083333	Sample Variance	356.8095238
Kurtosis	-0.328200158	Kurtosis	-0.881402202
Skewness	0.308984848	Skewness	-0.242355616
Range	113	Range	52
Minimum	403	Minimum	362
Maximum	516	Maximum	414
Sum	5443	Sum	2738
Count	12	Count	7
Confidence Level(80.0%)	12.95906469	Confidence Level(80.0%)	10.27916932
SV=SAMPLE VARIANCE
		11* sv(A) =	11924.91667
df= 12+7-2 =17		6* sv(B)=	2140.857143
		sum=	14065.77381
Poold variance S2 =	1/ (12+7 -2) {11*sv(A)+6*sv(B)}
	827.3984594
S.E. (mean1-mean2)=	sqrt{S2(1/12+1/8)} =
	sqrt{827.39846* 0.22619}=
	sqrt{ 187.1496515)=
	13.680265
80% confidence interval=
(mean1 - mean2) ±	t0.20 *(S.E.)
62.44047619 ±	1.333*S.E.
62.44047619 ±	18.23579325
LB=	44.20468295
UB=	80.67626944
	ANSWER