a)A regeneration survey has indicated that 85 of the 110 plots examined were sto

Question

a)A regeneration survey has indicated that 85 of the 110 plots examined were stocked. If the requirement is 80% stocking for satisfactory regeneration, would you call this area as "satisfactory"? Use = 0.01. b)Another survey from another area indicated 84% stocking from 100 plots. Is this area different from the area described in Question (a)? Use = 0.05. If the germination capacity of a seed lot is at least 70%, it is acceptable for a nursery. A test of a given seed lot showed that 8 out of 18 seeds germinated. Would you accept this seed lot? Use = 0.05 (or as close as possible).

Explanation / Answer

Answer to part a)

Sample proportion P^ = 85/110 = 0.7727

Population Proportion P = 0.80

n = 110

.

SE = sqrt(P*Q/n)

SE = sqrt(0.80 *0.20 /110)

SE = 0.03814

.

Formula of Test Statistic

Z = (P^-P) / SE

Z = (0.7727 - 0.80) / 0.03814

Z = -0.72

.

The P value for this is : 0.2358

.

Since the P value is greater than alpha 0.01 , we conclude this area to be satisfactory

.

Answer to part b)

Z = (0.84 - 0.80) / 0.03814

Z = 1.05

.

The P value is 0.1469

.

Since the P value is greater than alpha 0.05 , we conclude this area to be satisfactory

.

Answer to part c)

The P^ = 8/18 = 0.44

P = 0.70

n = 18

.

SE = sqrt(0.70*0.30/18)

SE = 0.1080

.

Z = (0.4444 - 0.70) / 0.1080

Z = -2.37

.

The P value is : 0.0089

.

Since the P value 0.0089 < 0.05 , we fail to accept this lot

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