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My Notes Ask Your Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ-7. The hypotheses H0: μ 74 and Hai μ < 74 are to be tested using a random sample of n25 observations. (a) How many standard deviations (of X) below the null value is x 72.3? (Round your answer to two decimal places.) 1 standard deviations (b) If x72.3, what is the conclusion using a 0.002 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.) 28782x -value . 0.113, State the conclusion in the problem context. O Reject the null hypothesis. There is not sufficient evidence to O Reject the null hypothesis. There is sufficient evidence to conclude O Do not reject the null hypothesis. There is sufficient evidence to o Do not reject the null hypothesis. There is not sufficient evidence conclude that the mean drying time is less than 74 that the mean drying time is less than 74 conclude that the mean drying time is less than 74. to conclude that the mean drying time is less than 74 (c) For the test procedure with a 0.002, what is B(70)? (Round your answer to four decimal places.) (d) If the test procedure with α-0.002 is used, what n is necessary to ensure that B(70) 0.01? (Round your answer up to the next whole number.) n6x specimens (e) If a level 0.01 test is used with n 100, what is the probability of a type I error when 76? (Round your answer to four decimal places.) 05080 x You may need to use the appropriate table in the Appendix of Tables to answer this question.
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Answer #1

b)

z =-1.21

c)

Hypothesized mean true mean standard deviation sample size standard error of mean for 0.002 level and left tailed test critival value Ζα- rejection region: X <-H+Zaox or type li error-probability of not rejecting β P(Z >(69.9706-μα)/OX)) 74 70 7 25 1.4000 2.88 69.9706 FHa En P(Z >-0.02) 0.5080

d)

e)

P(type I error )=P(Xbar <72.37)=P(Z<(72.37-76)*sqrt(100)/7)=P(Z<-5.19)=0.0000

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