1. The life in hours of a thermocouple used in a furnace is known to be approxim

Question

1. 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)(5 pts) Calculate the estimator of the mean life time (b)(5 pts) Construct a 95% confidence interval of the mean life time (c)(5 pts) Construct a 95% prediction interval of the mean life time (d)(5 pts) Is there evidence to support the claim that mean life exceeds 540 hours? State the null hypothesis and alternative hypothesis. (e) (5 pts) Calculate the standarized test statics (f) (5 pts) What is your conclusion? Is the n hypothesis rejected or not?

Explanation / Answer

1.

a.
mean life time = sum of observations/number of observations
= 553+552+567+579+550+541+537+553+552+546+538+553+581+539+529/15
= 551.3333
b.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =551.33
population size (n)=15
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 15) )
= 5.164
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
value of z table is 1.64
margin of error = 1.64 * 5.164
= 8.469
III.
CI = x ± margin of error
confidence interval = [ 551.33 ± 8.469 ]
= [ 542.861,559.799 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =551.33
population size (n)=15
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
value of z table is 1.64
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 = [ 551.33 ± Z a/2 ( 20/ Sqrt ( 15) ) ]
= [ 551.33 - 1.64 * (5.164) , 551.33 + 1.64 * (5.164) ]
= [ 542.861,559.799 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [542.861 , 559.799 ] 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 = 551.33
standard error =5.164
z table value = 1.64
margin of error = 8.469
confidence interval = [ 542.861 , 559.799 ]
c.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =551.33
population size (n)=15
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 15) )
= 5.164
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.025
from standard normal table,left tailed z /2 =1.96
since our test is left-tailed
value of z table is 1.96
margin of error = 1.96 * 5.164
= 10.121
III.
CI = x ± margin of error
confidence interval = [ 551.33 ± 10.121 ]
= [ 541.209,561.451 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =551.33
population size (n)=15
level of significance, = 0.025
from standard normal table,left tailed z /2 =1.96
since our test is left-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 = [ 551.33 ± Z a/2 ( 20/ Sqrt ( 15) ) ]
= [ 551.33 - 1.96 * (5.164) , 551.33 + 1.96 * (5.164) ]
= [ 541.209,561.451 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 97.5% sure that the interval [541.209 , 561.451 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 97.5% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 551.33
standard error =5.164
z table value = 1.96
margin of error = 10.121
confidence interval = [ 541.209 , 561.451 ]
d.
Given that,
population mean(u)=540
standard deviation, =20
sample mean, x =551.33
number (n)=15
null, Ho: =540
alternate, H1: >540
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 551.33-540/(20/sqrt(15)
zo = 2.19405
| zo | = 2.19405
critical value
the value of |z | at los 5% is 1.645
we got |zo| =2.19405 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.19405 ) = 0.01412
hence value of p0.05 > 0.01412, here we reject Ho
ANSWERS
---------------
null, Ho: =540
alternate, H1: >540
e.
test statistic: 2.19405
critical value: 1.645
f.
decision: reject Ho
p-value: 0.01412
we have enough evidence to support the claim

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