MULTIPLE CHOICE. Choose the one alternative that best completes the statement or

Question

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the P-value for the indicated hypothesis test. 18 An airline claims that the no-show rate for passengers booked on its flights is less than 6% of 380 randomly selected reservations, 18 were no-shows. Find the P-value for a test of the airline's claim. 18) A) 0.3508 B) 0.1492 0.1230 D) 0.0746 19) Find the P-value for a test of the claim that less than 50% of the people following a particular diet 19) will experience increased energy. Of 100 randomly selected subjects who followed the diet, 47 noticed an increase in their energy level. A) 0.2257 B) 0.5486 C) 0.2743 D) 0.7257

Explanation / Answer

1) One proportion test

H0 : p= 6%=0.06 Vs Ha : p < 6% =0.06

sample proportion: 18/380 = 0.047

Test statistics :

Z = (p - p0) / sqrt( (p0 (1 - p0)/n )) = (0.047 -0.06)/ sqrt( (0.06*(1-0.06))/ 380) = -1.03987

Test is left tailed because alternative hypothesis contains < sign.

P( Z< -1.03987) = 0.1492

By using excel command  

type in any emppty excel cell

=NORMSDIST(-1.03987) and hit the enter it will gives P(Z< -1.03987) = 0.1492

So correct option is B

2)

H0 : p= 0.50 Vs Ha : p < 0.50

sample proportion: 47/100 = 0.47

Test statistics :

Z = (p - p0) / sqrt( (p0 (1 - p0)/n )) = (0.47 -0.5)/ sqrt( (0.5*(1-0.5))/ 100) = -0.6

Test is left tailed because alternative hypothesis contains < sign.

P( Z< -0.6) = 0.2743

By using excel command  

type in any emppty excel cell

=NORMSDIST(-0.6) and hit the enter it will gives P(Z< -0.6) = 0.2743

So correct option is C

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