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marshall27 [118]
3 years ago
15

The table below gives selected entries from the 2014 federal income tax table. It applies to a married couple filing jointly. Ta

xable income and tax owed are both in dollars.
Taxable
income 73,300 73,500 73,700 73,900 74,100 74,300
Tax owed 10,091 10,121 10,151 10,194 10,244 10,294
Taxable
income 74,500 74,700 74,900 75,100 75,300 75,500
Tax owed 10,344 10,394 10,444 10,494 10,544 10,594
Make a table that shows the additional tax owed over each income span.
Taxable
income 73,300 73,500 73,700 73,900 74,100 74,300
Tax owed 10,091 10,121 10,151 10,194 10,244 10,294
Additional
tax ($)
Taxable
income 74,500 74,700 74,900 75,100 75,300 75,500
Tax owed 10,344 10,394 10,444 10,494 10,544 10,594
Additional
tax ($)

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0

Step-by-step explanation:

first think is income,taxable charge ,taxable yearly

charge

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6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=6.26

The sample deviation calculated s=2.47

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Assuming a Confidence of 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

Now we have everything in order to replace into formula (1):

6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)    

6 0
3 years ago
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