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marissa [1.9K]
3 years ago
15

Bethany uses 8% of her paycheck to pay for gas. If she spent 34.12 on gas this week how much was her paycheck?

Mathematics
1 answer:
Schach [20]3 years ago
7 0

Answer: 426.5

Step-by-step explanation:

You know that she spent 34.12 on gas and you also know that that amount of money was the 8% of her paycheck.

Therefore, keeping the above on mind, you can write the following expression, where x is the total amount of money of her paycheck:

0.08x=34.12

When you solve for  x, you obtain the result shown below:

x=\frac{34.12}{0.08}\\\\x=426.5

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Answer:

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

p_v =P(t_{(5)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

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Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

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

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

\bar X=29.35 represent the sample mean  

s=1.365 represent the sample standard deviation  

n=6 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :  

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Alternative hypothesis:\mu < 30  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

P-value  

We need to calculate the degrees of freedom first given by:  

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Since is a one-side left tailed test the p value would given by:  

p_v =P(t_{(5)}  

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