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inna [77]
3 years ago
12

Suppose a college student pays $750 for tuition fees. However, she also has to pay $300 for her textbooks (ouch!). What percent

of her total education costs does she pay for her books?
Mathematics
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

Total costs = $700 + $300 = $1000.

 $300 / $1000 = 0.3 = 3%

Step-by-step explanation:

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Isn’t it 0.44444444.....or 0.444
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Answer: it is 100% A

Step-by-step explanation:

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3x11111111111111111111111111
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3x1=4

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Which expression is equivalent to √120x<br> 2√30x<br> 2√30<br> 4√30x<br> 60√2x
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A well-known brokerage firm executive claimed that 50% of investors are currently confident of meeting their investment goals. A
VMariaS [17]

Answer:

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.005 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 adults that said they are confident of meeting their goals  is significantly lower than 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X represent the people that said they are confident of meeting their goals

\hat p=0.44 estimated proportion of adults that said they are confident of meeting their goals

p_o=0.5 is the value that we want to test

\alpha=0.005 represent the significance level

Confidence=99.5% or 0.995

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.5.:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

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.

3) Calculate the statistic  

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

z=\frac{0.44 -0.5}{\sqrt{\frac{0.5(1-0.5)}{200}}}=-1.697  

4) 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=0.005. The next step would be calculate the p value for this test.  

Since is left tailed test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.005 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 adults that said they are confident of meeting their goals  is significantly lower than 0.5 or 50% .  

4 0
3 years ago
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