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hw6-2 Sol

This document provides the solution to a statistics homework problem involving hypothesis testing of a population proportion. It tests whether the proportion of automobile engine crankshaft bearings with excess surface roughness exceeds 0.06 based on a sample of 85 bearings where 7 had roughness that exceeded specifications. The null and alternative hypotheses are stated. The test statistic is calculated and the p-value is found to be 0.1977 which is greater than the significance level of 0.1, so the null hypothesis is not rejected. A 90% confidence interval for the population proportion is also calculated and found to include the value of 0.06 in the null hypothesis, consistent with failing to reject the null hypothesis.

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52 views1 page

hw6-2 Sol

This document provides the solution to a statistics homework problem involving hypothesis testing of a population proportion. It tests whether the proportion of automobile engine crankshaft bearings with excess surface roughness exceeds 0.06 based on a sample of 85 bearings where 7 had roughness that exceeded specifications. The null and alternative hypotheses are stated. The test statistic is calculated and the p-value is found to be 0.1977 which is greater than the significance level of 0.1, so the null hypothesis is not rejected. A 90% confidence interval for the population proportion is also calculated and found to include the value of 0.06 in the null hypothesis, consistent with failing to reject the null hypothesis.

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STAT 509 2016 Fall HW6-2 Solution

Instructor: Shiwen Shen


Collection Day: October 5

1. In a random sample of 85 automobile engine crankshaft bearings, 7 have a surface finish rough-
ness that exceeds the specifications. Does this data present sufficient evidence that the pro-
portion of crankshaft bearings, say p, exhibiting excess surface roughness is greater than 0.06?
We will address this using a hypothesis test.
(a) State the null and alternative hypotheses.

H0 : p = 0.06
Ha : p > 0.06

(b) Calculate the appropriate test statistic.


7
We know the sample proportion is p̂ = 85 = 0.082, so the test statistic is

p̂ − p0 0.082 − 0.06
z0 = q = q = 0.85
p0 (1−p0 ) 0.06×0.94
n 85

(c) What is the p-value of the test?


Given Ha : p > 0.06, we have

p-value = P (Z > z0 ) = P (Z > 0.854) = 1 − P (Z < 0.85) = 1 − 0.8023 = 0.1977

(d) What is your conclusion based on the p-value if we use α = 0.1?


Because p-value= 0.1977 > 0.1 = α, we fail to reject the null hypothesis H0 . Therefore,
with 90% confidence level, we fail to conclude that the proportion of crankshaft bearings
exhibiting excess surface roughness is greater than 0.06.
(e) Calculate a 90% confidence interval for the population proportion. Interpret the confidence
interval using the context of the question.
r
p̂(1 − p̂)
C.I. = p̂ ± zα/2
rn
0.082(1 − 0.082)
= 0.082 ± 1.64
85
= 0.082 ± 0.058
= (0.024, 0.14)

(f) Compare the results in (d) and (e), are they similar?
0.06 is included in the 90% confidence interval (0.024, 0.14). It is consistent to the conclu-
sion that we fail to reject H0 : p = 0.06.

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