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SOLVE THIS PROBLEM BY USING SPSS PROGRAM Two varieties of wheat were tested usin

ID: 3049291 • Letter: S

Question

SOLVE THIS PROBLEM BY USING SPSS PROGRAM

Two varieties of wheat were tested using two similar plots on eight different farms. One plot on each farm was selected at random to receive variety A, and the other variety B. All plots were planted on the same day and managed identically. The yields (kg/plot) 1 2345 6 7 8 Farm Variety A 17.8 18.5 12.2 19.7 10.8 11.9 15.6 12.5 Variety B 14.7 15.2 12.9 18.3 10.1 12.2 13.5 9.9 Is there evidence at the 5% significance level that the yield of the 2 varieties differ? a) b) Estimate the difference in yield with 95% confidence.

Explanation / Answer

Part (a)

Let X = yield from variety A

      Y = yield from variety B

Then, X ~ N(µ1, 12) and Y ~ N(µ2, 22), where 12 = 22 = 2, say and 2 is unknown.

Claim:

The yield of two varieties differ.

Hypotheses:

Null: H0: µ1 = µ2 Vs Alternative: HA: µ1 µ2

Test Statistic:

t = (Xbar - Ybar)/{s(2/n)} where

s2 = (s12 + s22)/2;

Xbar and Ybar are sample averages and s1, s2 are sample standard deviations based on n observations each on X and Y.

Calculations

Summary of Excel calculations is given below:

n =

8

Xbar =

14.875

Ybar =

13.35

s1 =

3.459046

s2 =

2.771281

s^2 =

9.8225

s =

3.134087

tcal =

0.97317

=

0.05

tcrit =

2.144787

p-value =

0.346984

Distribution, Critical Value and p-value:

Under H0, t ~ t2n - 2. Hence, for level of significance %, Critical Value = upper (/2)% point of t2n - 2 and p-value = P(t2n - 2 > | tcal |).

Using Excel Functions, the above are found to be as follows:

tcrit =

2.144787

p-value =

0.346984

Decision Criterion (Rejection Region):

Reject H0 if | tcal | > tcrit or p-value <

Decision:

Since | tcal | < tcrit, H0 is accepted. Since p-value > , H0 is accepted.

Conclusion:

There is not sufficient evidence to suggest that the claim is valid.

=> the two varieties do not differ with respect to yield.

DONE

Part (b)

100(1 - ) % Confidence Interval for (1 - 2) is: (Xbar – Ybar) ± {(t2n – 2, /2)(s)(2/n)}

= 1.525 ± (2.1448 x 3.1341 x 0.5)

= 1.525 ± 3.361

= (1.836, 4.886) ANSWER

n =

8

Xbar =

14.875

Ybar =

13.35

s1 =

3.459046

s2 =

2.771281

s^2 =

9.8225

s =

3.134087

tcal =

0.97317

=

0.05

tcrit =

2.144787

p-value =

0.346984