The quality-control manager at a compact fluorescent light bulb (CFL) factory ne

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The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to.....

The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7,495 hours. The population standard deviation is 700 hours. A random sample of 49 light bulbs indicates a sample mean life of 7,345 hours a. At the 0.05 level of significance is there evidence that the mean life is different from 7,495 hours? b. Compute the p-value and interpret its meaning C. Construct a 95% confidence interval estimate of the population mean life ofthe light bulbs. d. Compare the results of (a) and (c). What conclusions do you reach? a. Let be the population mean Determine the null hypothesis, H0, and the alternative hypothesis H1 d. Compare the results of (a) and (c). What conclusions do you reach? O A. The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7,495 hours ° C. The results of (a) and (c) are not the same. there is sufficient evidence to prove that the mean life is different from 7,495 hours. What is the test statistic? ZSTAT(Round to two decimal places as needed)OC. The results of (a) and (e) are not the same there is sufficient evidence to prove that the mean life is different from 7,495 hours O D. The results of (a) and (c) are the same, there is sufficient evidence to prove that the mean life is different from 7.495 hours. What is are the critical value(s)? Round to two decimal places as needed. Use a comma to separate answers as needed.) What is the final conclusion? O A. Reject Ho. There is sufficient evidence to prove that the mean life is different from 7,495 hours O B. Reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,495 hours OC. Fail to reject Ho There is sufficient evidence to prove that the mean life is different from 7,495 hours. D. Fail to reject Ho . There is not sufficient evidence to prove that the mean life is different from 7.495 hours. b. What is the p-value? Round to three decimal places as needed.) Interpret the meaning of the p-value. Choose the correct answer below ( A. Fail to reject Ho . There is sufficient evidence to prove that the mean life is different from 7495 hours. B. Fail to reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,495 hours. O C. Reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,495 hours. 0 D. Reject Ho. There is sufficient evidence to prove that the mean life is different from 7,495 hours. C. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. sps (Round to one decimal place as needed )

Explanation / Answer

Given that,
population mean(u)=7495
standard deviation, =700
sample mean, x =7345
number (n)=49
null, Ho: =7495
alternate, H1: !=7495
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 7345-7495/(700/sqrt(49)
zo = -1.5
| zo | = 1.5
critical value
the value of |z | at los 5% is 1.96
we got |zo| =1.5 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.5 ) = 0.13
hence value of p0.05 < 0.13, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: =7495
alternate, H1: !=7495
test statistic: -1.5
critical value: -1.96 , 1.96
decision: do not reject Ho
Fail to reject null hypothesis.there is sufficient evidence to prove that the mean life is different from 7495 hours
b.
p-value: 0.13
Fail to reject null hypothesis.there is sufficient evidence to prove that the mean life is different from 7495 hours
c.
TRADITIONAL METHOD
given that,
standard deviation, =700
sample mean, x =7495
population size (n)=49
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 700/ sqrt ( 49) )
= 100
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 100
= 196
III.
CI = x ± margin of error
confidence interval = [ 7495 ± 196 ]
= [ 7299,7691 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =700
sample mean, x =7495
population size (n)=49
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 7495 ± Z a/2 ( 700/ Sqrt ( 49) ) ]
= [ 7495 - 1.96 * (100) , 7495 + 1.96 * (100) ]
= [ 7299,7691 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [7299 , 7691 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 7495
standard error =100
z table value = 1.96
margin of error = 196
confidence interval = [ 7299.0 , 7691.0 ]
d.
part(a) and (c) are same but there is not sufficient evidence to prove that mean life different from 7495 hours

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