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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Audrey 1 month 2021-10-18T13:01:02+00:00 1 Answer 0 views 0

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    2021-10-18T13:03:00+00:00

    Answer:

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

    p_v =P(z>2.063)=0.020  

    So the p value obtained was a low value and using the significance level given \alpha=0.05 we have p_v<\alpha 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(z>2.06)=0.020  

    So the p value obtained was a low value and using the significance level given \alpha=0.05 we have p_v<\alpha 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.

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