Let x be a random variable that represents assembly times for the Ford Taurus. A
ID: 3231866 • Letter: L
Question
Let x be a random variable that represents assembly times for the Ford Taurus. A research journal reported that the typical assembly time had been 38 hours. A modification to the assembly procedure has been made. Experience with assembly methods indicates that = 1.8 hours. It is thought that the average assembly time may be reduced by this new modification. A random sample of 50 new Ford Taurus automobiles coming off the assembly line showed the average assembly time using the new method to be x = 37.74 hours. Does the data support the claim that the average assembly time has now been reduced? Use = 0.10.
(a) Identify the claim, the null hypothesis, and the alternative hypothesis.
(b) Will you use a left-tailed, right-tailed, or two-tailed test?
two-tailed
left-tailed
right-tailed
The Student's t, since the sample size is large and is known.
The standard normal, since the sample size is large and is unknown.
The standard normal, since we know that x has a normal distribution with known .
The standard normal, since the sample size is large and is known.
The standard normal, since we know that x has a normal distribution with unknown .
The Student's t, since the sample size is large and is unknown.
At the = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.
At the = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
Reject the null hypothesis, there is sufficient evidence that the average assembly time is less than 38 hours.
Fail to reject the null hypothesis, there is insufficient evidence that the average assembly time is less than 38 hours.
Fail to reject the null hypothesis, there is sufficient evidence that the average assembly time is less than 38 hours.
Reject the null hypothesis, there is insufficient evidence that the average assembly time is less than 38 hours.
Claim: > < = 38 Ho: > < = 38H1: > < = 38
(b) Will you use a left-tailed, right-tailed, or two-tailed test?
two-tailed
left-tailed
right-tailed
(c) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The Student's t, since the sample size is large and is known.
The standard normal, since the sample size is large and is unknown.
The standard normal, since we know that x has a normal distribution with known .
The standard normal, since the sample size is large and is known.
The standard normal, since we know that x has a normal distribution with unknown .
The Student's t, since the sample size is large and is unknown.
(d) Sketch the sampling distribution showing the area corresponding to the approximate P-value.
(e) Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?
At the = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.
At the = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(f) State your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the average assembly time is less than 38 hours.
Fail to reject the null hypothesis, there is insufficient evidence that the average assembly time is less than 38 hours.
Fail to reject the null hypothesis, there is sufficient evidence that the average assembly time is less than 38 hours.
Reject the null hypothesis, there is insufficient evidence that the average assembly time is less than 38 hours.
Explanation / Answer
Given that,
population mean(u)=38
standard deviation, =1.8
sample mean, x =37.74
number (n)=50
null, Ho: =38
alternate, that the average assembly time may be reduced by this new modification, H1: <38
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.282
since our test is left-tailed
reject Ho, if zo < -1.282
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 37.74-38/(1.8/sqrt(50)
zo = -1.02138
| zo | = 1.02138
critical value
the value of |z | at los 10% is 1.282
we got |zo| =1.02138 & | z | = 1.282
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.02138 ) = 0.15354
hence value of p0.1 < 0.15354, here we do not reject Ho
ANSWERS
---------------
null, Ho: =38
alternate, H1: <38
test statistic: -1.02138
critical value: -1.282
decision: do not reject Ho
p-value: 0.15354
left-tailed
The standard normal, since the sample size is large and is known.
Fail to reject the null hypothesis, there is insufficient evidence that the average assembly time is less than 38 hours.
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