Mathematics, 10.12.2019 05:31, mathwhiz9125

Arecent article reported that a job awaits only one in three new college graduates. the major reasons given were an overabundance of college graduates and a weak economy. a survey of 200 recent graduates from your school revealed that 80 students had jobs. at the 0.01 significance level, can we conclude that a larger proportion of students at your school have jobs?

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answered: teamroper35

$z=\frac{0.4 -0.33}{\sqrt{\frac{0.33(1-0.33)}{200}}}=2.105$

p_v =P(z2.105)=0.018

If we compare the p value obtained and the significance level given $\alpha=0.01$ we have $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of students that had job at the school mentioned is not significantly higher then 0.33 .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=80 represent the number of students that had jobs

$\hat p=\frac{80}{200}=0.4$ estimated proportion of students that had jobs

p_o=0.3333 is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 of students that have jobs at the school mentioned is higher than the reported value at the article:

Null hypothesis: $p\leq 0.33$

Alternative hypothesis: p 0.33

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:

$z=\frac{0.4 -0.33}{\sqrt{\frac{0.33(1-0.33)}{200}}}=2.105$

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 $\alpha=0.01$ . 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(z2.105)=0.018

answered: Guest

the answer for this question is 38

answered: Guest

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: