Question

In a metal fabrication process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are normally distributed. A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet. Based on this information answer the following questions to determine if the manufacturer is making rods too short.

a. Write the null and alternative hypotheses to test this claim.

b. What is the value of the test statistic? You must write down the formula to get full credit.

c. What is the associated P-value?

d. State your conclusion and explain in context using ΅ = 0.05.

e. Create a 95% Confidence Interval for the true mean length of the rods. Explain what the confidence interval means in context. For example..."We are 95% confident..."

f. Extra Credit: Create a 95% Confidence Interval for the population Standard Deviation of the rods.

Answer 1

Given that,
population mean(u)=15
sample mean, x =14.8
standard deviation, s =0.65
number (n)=16
null, Ho: μ=15
alternate, H1: μ!=15
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.131
since our test is two-tailed
reject Ho, if to < -2.131 OR if to > 2.131
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =14.8-15/(0.65/sqrt(16))
to =-1.2308
| to | =1.2308
critical value
the value of |t α| with n-1 = 15 d.f is 2.131
we got |to| =1.2308 & | t α | =2.131
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.2308 ) = 0.2374
hence value of p0.05 < 0.2374,here we do not reject Ho
ANSWERS
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a.
null, Ho: μ=15
alternate, H1: μ!=15
b.
test statistic: -1.2308
critical value: -2.131 , 2.131
decision: do not reject Ho
c.
p-value: 0.2374
d.
we do not have enough evidence to support the claim that metal fabrication process, metal rods are produced to a specified target length of 15 feet.
e.
TRADITIONAL METHOD
given that,
sample mean, x =14.8
standard deviation, s =0.65
sample size, n =16
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.65/ sqrt ( 16) )
= 0.163
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
margin of error = 2.131 * 0.163
= 0.346
III.
CI = x ± margin of error
confidence interval = [ 14.8 ± 0.346 ]
= [ 14.454 , 15.146 ]
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DIRECT METHOD
given that,
sample mean, x =14.8
standard deviation, s =0.65
sample size, n =16
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 14.8 ± t a/2 ( 0.65/ Sqrt ( 16) ]
= [ 14.8-(2.131 * 0.163) , 14.8+(2.131 * 0.163) ]
= [ 14.454 , 15.146 ]
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interpretations:
1) we are 95% sure that the interval [ 14.454 , 15.146 ] 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
f.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.05
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.025 = 0.975
the two critical values ᴪ^2 left, ᴪ^2 right at 15 df are 27.4884 , 6.262
s.d( s )=0.65
sample size(n)=16
confidence interval for σ^2= [ 15 * 0.4225/27.4884 < σ^2 < 15 * 0.4225/6.262 ]
= [ 6.3375/27.4884 < σ^2 < 6.3375/6.2621 ]
[ 0.2306 < σ^2 < 1.012 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (0.2306) < σ < sqrt(1.012), ]
= [ 0.4802 < σ < 1.006 ]