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- Homework Statement
- An audit analysis shows that over the last ##6## years, the mean amount per sales invoice for customers has been ##$110##. A random sample of ##12## sales invoices was selected.

The Auditor is interested in whether or not there is evidence of a change in the mean amount of sales invoice from the ##$110## amount at ##5\%## level of significance.

note; i was unable to copy the table having the data on this page...

- Relevant Equations
- stats; see question below;

Consider the given table below;

My take;

We shall use t-distribution as the given sample is less than ##30##. Therefore using ##(n-1)## degree of freedom and ##95\%## confidence level we shall have ##t_{statistic}= 2.2010##

Now on the calculation bit;

We have ##t_{calculated} =\dfrac{x-μ}{\dfrac{s}{\sqrt{n}}}##

From my calculation i have;

sample standard deviation=##\sqrt{\dfrac{4758.1076}{11}}=\sqrt{432.555}=20.8##

sample mean=## \dfrac{1366.33}{12}=113.86##

##t_{calculated} =\dfrac{113.86-110}{\dfrac{20.8}{\sqrt{12}}}=\dfrac{3.86}{6}=0.643##since, ##t_{statistic} >t_{calculated} ##, then we can conclude that there is change in the mean amount of sale invoices.

I have no solution to the problem. Your insight highly welcome.

109.99 | 112.46 | 153.23 | 111.60 | 128.47 | 108.27 |

94.33 | 92.98 | 112.57 | 76.72 | 129.59 | 136.12 |

We shall use t-distribution as the given sample is less than ##30##. Therefore using ##(n-1)## degree of freedom and ##95\%## confidence level we shall have ##t_{statistic}= 2.2010##

Now on the calculation bit;

We have ##t_{calculated} =\dfrac{x-μ}{\dfrac{s}{\sqrt{n}}}##

From my calculation i have;

sample standard deviation=##\sqrt{\dfrac{4758.1076}{11}}=\sqrt{432.555}=20.8##

sample mean=## \dfrac{1366.33}{12}=113.86##

##t_{calculated} =\dfrac{113.86-110}{\dfrac{20.8}{\sqrt{12}}}=\dfrac{3.86}{6}=0.643##since, ##t_{statistic} >t_{calculated} ##, then we can conclude that there is change in the mean amount of sale invoices.

I have no solution to the problem. Your insight highly welcome.

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