# Solve the given problem that involves Test of Hypothesis

• chwala
In summary: And be sure to state the degrees of freedom, the critical value, and the test statistic. In summary, 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, using a two-tailed test with 11 degrees of freedom and a critical value of 2.2010. Based on the sample data, the test statistic is 0.643 and the conclusion is to reject the null hypothesis and accept the alternative hypothesis that there is a change in the mean amount of the sales invoices. #### chwala Gold Member 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. Last edited: 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.

chwala

## 1. What is a test of hypothesis?

A test of hypothesis is a statistical method used to determine whether a hypothesis about a population is supported by the data. It involves collecting and analyzing data to determine the likelihood of the observed results occurring by chance.

## 2. How is a test of hypothesis conducted?

A test of hypothesis involves several steps. First, a null hypothesis is established, which is the assumption that there is no significant difference or relationship between variables. Then, data is collected and analyzed using statistical methods to determine the probability of the observed results occurring if the null hypothesis is true. If the probability is low enough, the null hypothesis is rejected in favor of an alternative hypothesis.

## 3. What is the significance level in a test of hypothesis?

The significance level, also known as alpha, is the predetermined threshold used to determine whether the null hypothesis should be rejected. It is typically set at 5% or 0.05, meaning that if the probability of the observed results occurring by chance is less than 5%, the null hypothesis is rejected.

## 4. What is a type I error in a test of hypothesis?

A type I error occurs when the null hypothesis is rejected even though it is actually true. This means that the results of the test incorrectly lead to the conclusion that there is a significant difference or relationship between variables when there is not.

## 5. What is a type II error in a test of hypothesis?

A type II error occurs when the null hypothesis is not rejected even though it is actually false. This means that the results of the test incorrectly lead to the conclusion that there is no significant difference or relationship between variables when there actually is.