Please what is the answer?

Katena32 [7]

I’m solving for x
X =2/3

5 0

3 years ago

A recent article reported that a job awaits only one in three new college graduates. The major reasons given were an overabundan

nikitadnepr [17]

Answer:

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

$p_v =P(z>2.105)=0.018$

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:

$p_v =P(z>2.105)=0.018$

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 .

7 0

3 years ago

What is the slope of a line perpendicular to the line whose equation is 3x+y=-5

julsineya [31]

Answer:

tdy

Step-by-step explanation:

tyzey

7 0

3 years ago

A certain number divided by 4 equals 6. find the number let n= the number

AURORKA [14]

24 dvided by 4 is equal to 6. :)

7 0

3 years ago

How are natural numbers, whole numbers and integers used in everyday life?

Nezavi [6.7K]

Natural numbers: Counting things! You look around your room and see an electronic device, then another, then another! You just counted to 3 using the natural numbers.

Whole numbers! You try to look for electronic devices and realise that they’re all gone. You have zero electronic devices, and you just used whole numbers.

You go online to find where your electronic devices went, and realise they were taken because you’re in debt to the bank so they took some of your stuff. You’re in negative numbers, and now you’ve used integers.

4 0

3 years ago