The life in hours of a thermocouple used in a furnace is known to be approximate

Question

The life in hours of a thermocouple used in a furnace is known to be approximately normally distributed with standard deviation = 20 hours. A random sample of 15 thermocouples resulted in the following data: 553, 552, 567, 579, 550, 541, 537, 553, 552, 546, 538, 553, 581, 539, 529. Hint: Z95% = 1:64, Z97:5% = 1:96. You need to determine which number to use in the following questions.

(a) Calculate the estimator of the mean life time

(b) Construct a 95% condence interval of the mean life time.

(c) Construct a 95% prediction interval of the mean life time.

(d) Is there evidence to support the claim that mean life exceeds 540 hours? State the null hypothesis and alternative hypothesis.

(e) Calculate the standarized test statics.

(f) What is your conclusion? Is the null hypothesis rejected or not?

Explanation / Answer

a)
Estimator of the mean life time = mean of given data = 551.2143

b)

d)Below are the null and alternate hypothesis

H0: mu = 540

H1: mu > 540

test statistics, t = (551.2143 - 540)/(14.8043/sqrt(15)) = 2.9338

p-value = 0.0054

f)

Reject null hypothesis as p-value is less than significance level of 0.05

CI for 95% n 15 mean 551.2142857 z-value of 95% CI 1.9600 std. dev. 14.8043361 SE = std.dev./sqrt(n) 3.82246 ME = z*SE 7.49189 Lower Limit = Mean - ME 543.72240 Upper Limit = Mean + ME 558.70618 95% CI (543.7224 , 558.7062 )

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