Brian says that 4×2/5=2/5+2/5+2/5. Describe and correct his error.

2 answers:

7 0

He is missing a 2/5. if he is multiplying it by 4 then when he is adding he need to add four of them together.

4x2/5= 1 and 3/5
2/5+2/5+2/5+2/5= 1 and 1/5
when he adds one more 2/5 he get 1 and 3/5

hope that helped!

gavmur [86]3 years ago

3 0

He did 4 and 2/5 multiples by 3 instead it should be 2/5x4

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Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t

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Answer:

Null hypothesis: $p \leq 0.69$

Alternative hypothesis: $p > 0.69$

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

$\hat p=0.71$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.69$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:

Null hypothesis: $p \leq 0.69$

Alternative hypothesis: $p > 0.69$

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:

$z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z>1.675)=0.047$

If we compare the p value obtained 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 adults that said that it is morally wrong to not report all income on tax returns is not significantly higher than 0.69.