Mathematics, 05.03.2020 17:34 caylah5921

Professor Jennings claims that only 35% of the students at Flora College work while attending school. Dean Renata thinks that the professor has underestimated the number of students with part-time or full-time jobs. A random sample of 83 students shows that 38 have jobs.
Do the data indicate that more than 35% of the students have jobs? Use a 5% level of significance.
What is the value of the sample test statistic? (Round your answer to two decimal places.)

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Answer from: brucewayne8499

$z=\frac{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06$

p_v =P(z2.063)=0.020

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ 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 students with jobs is significantly higher than 0.35.

Step-by-step explanation:

Data given and notation

n=83 represent the random sample taken

X=38 represent the students with jobs

$\hat p=\frac{38}{83}=0.458$ estimated proportion of students with jobs

p_o=0.35 is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 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:

$z=\frac{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06$

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 $\alpha=0.05$ . 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(z2.06)=0.020