## Professor Jennings claims that only 35% of the students at Flora College work while attending school. Dean Renata thinks that the professor

Question

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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1 month 2021-10-18T13:01:02+00:00 1 Answer 0 views 0  So the p value obtained was a low value and using the significance level given we have 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 estimated proportion of students with jobs is the value that we want to test represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest) 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: Alternative hypothesis: When we conduct a proportion test we need to use the z statistic, and the is given by: (1)

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this: 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 . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be: So the p value obtained was a low value and using the significance level given we have 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.