A sample of 14 joint specimens of a partieular type gave a sample mean proportio

Question

A sample of 14 joint specimens of a partieular type gave a sample mean proportional limit stress of 8.46 MPa and a sample standard deviation of 0.72 MPa. (a Caesate and interpret 9S% lower conndence bound nor the true average proportional int stress or all such oints. C Round your answer to two decimal places. MPa ? wth 95% conndence, we can say that the value of the true mean proportional int stress or all sich joints os greater than tNS value. with 9S% confidence, we can say that the value or the true mean proportional limit stress of all such joints is centered around this value. with 9S% confidence, we can say that the value of the true mean proportional limit stress of all such joints is less than this value. s did you make about the distribution of proportional limit stress O We must assume that the sample observations were taken from a normally distributed population. were taken from a uniformly distributed population We must assume that the sample We must assume that the sample observations were taken from a chi-square distributed population We do not need to make any assumptions. (D) Calculate and interpret a 95% lower prediction bound for proportional limit stress of 808 single joint of this type Round your answer to two decimal places. x MPa Interpret this bound.

Explanation / Answer

Part a

Answer:

Here, we have to find 95% confidence interval for the population mean. We are given

Sample mean = Xbar = 8.46

Sample standard deviation = S = 0.72

Sample size = n = 14

df = n – 1 = 14 – 1 = 13

Confidence level = 95%

Critical t value = 2.1604

(By using t table/excel)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 8.46 ± 2.1604*0.72/sqrt(14)

Confidence interval = 8.46 ± 2.1604* 0.192428094

Confidence interval = 8.46 ± 0.4157

Lower limit = 8.46 - 0.4157 = 8.0443

Upper limit = 8.46 + 0.4157 = 8.8757

Confidence interval = (8.04, 8.88)

Lower bound = 8.04

With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value.

We must assume the sample observations were taken from a normally distributed population.

Part b

Here, we have to find 95% lower prediction bound for proportional limit of a single joint of this type.

Lower prediction bound for single obs. = Xbar - t*S

Lower prediction bound = 8.46 - 2.1604*0.72

Lower prediction bound = 6.904512

Lower prediction bound = 6.90

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