Solve the given problem that involves Test of Hypothesis

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SUMMARY

The discussion centers on hypothesis testing using a sample of sales invoices to determine if there is a significant change in the mean amount from $110. The t-distribution is applied due to the sample size being less than 30, resulting in a t-statistic of 2.2010 and a calculated t-value of 0.643. The conclusion drawn is that since the t-statistic exceeds the calculated t-value, the null hypothesis is rejected, indicating a change in the mean amount of sales invoices. The importance of clearly stating null and alternative hypotheses in statistical testing is emphasized.

PREREQUISITES
  • Understanding of t-distribution and its application in hypothesis testing
  • Knowledge of calculating sample mean and standard deviation
  • Familiarity with null and alternative hypotheses in statistical analysis
  • Ability to interpret t-statistics and critical values
NEXT STEPS
  • Learn about the implications of one-tailed vs. two-tailed tests in hypothesis testing
  • Study the calculation and interpretation of confidence intervals
  • Explore the use of software tools like R or Python for statistical analysis
  • Investigate common pitfalls in hypothesis testing and how to avoid them
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Statisticians, data analysts, students in statistics courses, and professionals involved in data-driven decision-making will benefit from this discussion.

chwala
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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;

109.99​
112.46​
153.23​
111.60​
128.47​
108.27​
94.33​
92.98​
112.57​
76.72​
129.59​
136.12​
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.
 
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chwala said:
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;

Considdr the given table below;

109.99​
112.46​
153.23​
111.60​
128.47​
108.27​
94.33​
92.98​
112.57​
76.72​
129.59​
136.12​
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 no change in the mean amount of sale invoices.

I have no solution to the problem. Your insight highly welcome.
What are your null and alternate hypotheses? From the problem statement, "The Auditor is interested in whether or not there is evidence of a change in the mean amount of sales invoice." The "whether or not" part implies that the alternate hypothesis is ##H_a: \mu \ne 110##. If so, a two-tailed test would be appropriate. Would that make change in your calculations?
 
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Mark44 said:
What are your null and alternate hypotheses? From the problem statement, "The Auditor is interested in whether or not there is evidence of a change in the mean amount of sales invoice." The "whether or not" part implies that the alternate hypothesis is ##H_a: \mu \ne 110##. If so, a two-tailed test would be appropriate. Would that make change in your calculations?
Ok can i say,

Let the Null hypothesis = ##H_0=110##
Alternative hypothesis=##H_a≠110##

My calculations will still apply as shown in post ##1##.

...I end up with ##t_{statistic} [2.201]>t_{calculated} [0.643]## We therefore reject the Null hypothesis and accept the Alternative hypothesis; that there is change in the mean amount of the sales invoices.
 
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Looks good!

Tip: When you're doing a problem with inferential statistics, always include the null and alternate hypotheses. These hypotheses determine whether the critical region has one tail or two tails

Also, for these kinds of problems, be sure to state that the table numbers you're using are for the DoF, the confidence level, and whether you're using one tail or two tails.
 
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Mark44 said:
What are your null and alternate hypotheses? From the problem statement, "The Auditor is interested in whether or not there is evidence of a change in the mean amount of sales invoice." The "whether or not" part implies that the alternate hypothesis is ##H_a: \mu \ne 110##. If so, a two-tailed test would be appropriate. Would that make change in your calculations?
I think i made a mistake here,

Just by looking at the presented working, the t-statistic ought to be ##t=0.643## and the critical t-value ought to be ##t=2.201##

... another interesting aspect on hypothesis testing is that, the null hypothesis is never accepted. Correct? it is either we reject the null hypothesis or we fail to reject the null hypothesis.
 
chwala said:
... another interesting aspect on hypothesis testing is that, the null hypothesis is never accepted. Correct? it is either we reject the null hypothesis or we fail to reject the null hypothesis.
Correct? No. If we reject the alternate hypothesis, that's equivalent to accepting the null hypothesis.
 
Mark44 said:
Correct? No. If we reject the alternate hypothesis, that's equivalent to accepting the null hypothesis.
I just read on this, ... that in statistical terms, we never "accept" the null hypothesis; we only reject it or fail to reject it. This is because the absence of evidence against the null hypothesis does not prove it to be true; it just means we don't have sufficient evidence to show it’s false.

I will appreciate more insight on this.
 
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chwala said:
I just read on this, ... that in statistical terms, we never "accept" the null hypothesis; we only reject it or fail to reject it. This is because the absence of evidence against the null hypothesis does not prove it to be true; it just means we don't have sufficient evidence to show it’s false.

I will appreciate more insight on this.
Again, a failure to reject the null hypothesis is effectively a tacit acceptance of it.
 
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