Private and public universities are located in the same city. For the private university, 1042 alumni were surveyed and 655 said that they a

Question

Private and public universities are located in the same city. For the private university, 1042 alumni were surveyed and 655 said that they attended at least one class reunion. For the public university, 804 out of 1317 sampled alumni claimed they have attended at least one class reunion. Is the difference in the sample proportions statistically significant? (Use α=0.05)

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Samantha 1 week 2021-09-13T15:53:23+00:00 1 Answer 0

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    2021-09-13T15:55:08+00:00

    Answer:

    If we compare the p value with the significance level provided \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the true proportion’s for public and private universities are not significantly different at 5% of significance.  

    Step-by-step explanation:

    1) Data given and notation  

    X_{1}=655 represent the number of successes for private university

    X_{2}=804 represent the number of successes for public university

    n_{1}=1042 sample of 1 selected

    n_{2}=1317 sample of 2 selected

    \hat p_{1}=\frac{655}{1042}=0.629 represent the sample proportion for private university

    \hat p_{2}=\frac{804}{1317}=0.610 represent the sample proportion 2  

    z would represent the statistic (variable of interest)  

    p_v represent the value for the test (variable of interest)

    2) Concepts and formulas to use  

    We need to conduct a hypothesis in order to check if the we have significant differences betwen the two proportions, the system of hypothesis would be:  

    Null hypothesis:p_{1} = p_{2}  

    Alternative hypothesis:p_{1} \neq p_{2}  

    We need to apply a z test to compare proportions, and the statistic is given by:  

    z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)

    Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{655+804}{1042+1317}=0.618

    3) Calculate the statistic

    Replacing in formula (1) the values obtained we got this:  

    z=\frac{0.629-0.610}{\sqrt{0.618(1-0.618)(\frac{1}{1042}+\frac{1}{1317})}}=0.943  

    4) Statistical decision

    Since is a bilateral test the p value would be:  

    p_v =2*P(Z>0.943)=0.346  

    If we compare the p value with the significance level provided \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the true proportion’s for public and private universities are not significantly different at 5% of significance.  

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