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Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer decides to test this claim by sampling 35 parts. She finds that the standard deviation of the sample is 2.12. Is this enough evidence at the 1% level of significance, to accept the manufacturing companies claim? Calculate the appropriate test statistic (round to 2 decimal places as needed)

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Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to accept the manufacturing company claims

Explanation:

From the question we are told that

The sample size is n = 35

The sample standard deviation is
s =  2.12

The level of significance is
\alpha = 0.01

The null hypothesis is
H_o  :  \sigma ^2  = 9.0

The alternative hypothesis is
H_a &nbsp;: &nbsp;\sigma ^2 &nbsp;< &nbsp;9.0

Gnerally the test statistics is mathematically represented as


X^2 _(stat) = &nbsp;((n -1) * s^2 )/(\sigma^2 )


X^2 _(stat) = &nbsp;((35 &nbsp;-1) * 2.12^2 )/(9 )

=>
X^2 _(stat) = &nbsp;16.98

Generally the degree of freedom is mathematically represented as


df = &nbsp;n - 1

=>
df = &nbsp;35 - 1

=>
df = &nbsp;34

From the chi - distribution table the critical value of
\alpha at a degree of freedom of
df = &nbsp;34 is


X^2_(\alpha, 34 ) =56.060

From the value obtained we see that the test statistics does not lie within the region of rejection (1.e 56.060 ,
\infty )

Then

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to accept the manufacturing company claims