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Murljashka [212]
3 years ago
13

Company A has a sales position with a yearly salary of $42,000. Company B has a similar sales position with a slary of $39,000 p

lus 1% conmission on yearly sales. For what amount of yearly sales is the salary at company A greater than the salary and conmission at company B?
Mathematics
1 answer:
Ugo [173]3 years ago
8 0

Company A has a sales position with a yearly salary of $42,000. Company B has a similar sales position with a slary of $39,000 plus 1% conmission on yearly sales.

Let x be the amount of yearly sales

Company A has a sales position with a yearly salary of $42,000.

yearly salary = 42,000

Company B has a similar sales position with a salary of $39,000 plus 1% commission on yearly sales.

1% is 0.01

yearly salary = 39,000+ 0.01x

yearly sales is the salary at company A greater than the salary and conmission at company B

A > B

42000 > 39000+ 0.01x

We solve the inequality

39000+ 0.01x < 42000

Subtract 39000 on both sides

0.01x < 3000

divide by 0.01

x> 300,000

For yearly sales > $300,000, the salary at company A greater than the salary and commission at company B

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z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

\hat p=0.021 estimated proportion of interest

p_o=0.012 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:  

Null hypothesis:p \leq 0.012  

Alternative hypothesis:p > 0.012  

When we conduct a proportion test we need to use the z statisitic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

3 0
3 years ago
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