Question 260252: An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. Use a level of significance of = 0.01 to test the airlines claim. (Round phat to 4 decimal places.)
(References: examples 1 through 3 pages 408 - 410, end of section exercises 9 - 14 pages 411 - 412) (6 points)
1. H0 :
Ha :
2. =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:
6. Decision:
7. Interpretation:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. Use a level of significance of = 0.01 to test the airlines claim. (Round phat to 4 decimal places.)
(References: examples 1 through 3 pages 408 - 410, end of section exercises 9 - 14 pages 411 - 412) (6 points)
1.
H0 : p >= 0.05
Ha : p < 0.05 (airline claim)
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2. = 1%
3. Test statistics:
z(19/420) = (0.0452 - 0.05)= 0.0048/sqrt[0.05*0.95/420] = -0.4514
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4. P-value or critical z0 or t0.
p-value = P(z< -0.4478) = normalcdf(-100,-0.4514) = 0.3259..
----------------------------
5. Rejection Region:
cv = invNorm(0.01) = -2.326..
Reject Ho if ts < -2.326
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6. Decision: Since the p-value is greater than 1%, fail to reject Ho.
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7. Interpretation:
The proportion of no-shows is greater than or equal to 5% at
a 1% significance level. These test results support denying
the airline's claim.
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Cheers,
Stan H.
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