Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 8. The hypotheses H0: μ = 73 and Ha: μ < 73 are to be tested using a random sample of n = 25 observations.
(a) How many standard deviations (of X) below the null
value is x = 72.3? (Round your answer to two decimal
places.)
standard deviations
(b) If x = 72.3, what is the conclusion using α =
0.003?
Calculate the test statistic and determine the P-value.
(Round your test statistic to two decimal places and your
P-value to four decimal places.)
| z | = | |
| P-value | = |
State the conclusion in the problem context.
Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73. Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73.
(c) For the test procedure with α = 0.003, what is
β(70)? (Round your answer to four decimal places.)
β(70) =
(d) If the test procedure with α = 0.003 is used, what
n is necessary to ensure that β(70) = 0.01?
(Round your answer up to the next whole number.)
n = specimens
(e) If a level 0.01 test is used with n = 100, what is the
probability of a type I error when μ = 76? (Round your
answer to four decimal places.)
| a) number of standard deviation=(x̅-μ)*√n/σ = | 0.44 | ||||
| b) z score = | -0.44 | ||||
| p value = | 0.3300 | ||||
Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.
c)
| rejection region : x̅ <=μ+z*σ/√n = | 68.6000 | |||||
| β(70)=P(x̅ >68.6)=P(Z>(68.6-70)√25/8)=P(Z>-0.88)= | 0.8106 | |||||
d)
| required sample size n= | =(Zα+Zβ)2σ2/(μo-μa)2= | 184 | ||||
e)
| rejection region : x̅ <=μ+z*σ/√n = | 71.1360 | ||||||
| P(type I error)=P(x̅ <71.136)=P(Z<(71.136-76)√100/8)=P(Z<-6.08)= | 0.0000 | ||||||
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