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castortr0y [4]
3 years ago
6

You earn $28 for washing 7 cars. How much do you earn for washing 6 cars?

Mathematics
1 answer:
iren [92.7K]3 years ago
5 0

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

Since you can earn 28 dollars by washing 7 cars, and you want to find the amount money you can make by washing 6 cars.

\frac{28}{7} = \frac{x}{6}

First, you should find the amount of money you can make by washing 1 car (unit rate).

\frac{28}{7} = \frac{x}{1}

To get from 7 to 1, you have to divide 7 by 7; 7 ÷ 7 = 1.

Do the same thing to the numerator; 28 ÷ 7 = 4.

Now you know how much you make by washing 1 car, you need to find how much you make by washing 6 cars.

\frac{4}{1} = \frac{x}{6}

To get from 1 to 6, you multiply by 6. 1 times 6 equals 6.

Now do the same thing to the numerator; 4 × 6 = 24.

You can earn 24 dollars by washing 6 cars.

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

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Answer:

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

p_v =P(z>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

\hat p=\frac{67}{150}=0.447 estimated proportion of applicants who request financial aid

p_o=0.35 is the value that we want to test

\alpha represent the significance level

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 the true proportion of applicants who request financial aid is higher than 0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

When we conduct a proportion test we need to use the z statistic, 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{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

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 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>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

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