A researcher wants to check the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such

Question

A researcher wants to check the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds that the data was normally distributed and average jail time was 15.5 months with a standard deviation of 5.7 months. Is the reported claim correct? (use a 0.05 level of significance)

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Peyton 2 months 2021-10-09T03:21:54+00:00 1 Answer 0 views 0

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    2021-10-09T03:23:38+00:00

    Answer:

    t=\frac{15.5-18.7}{\frac{5.7}{\sqrt{11}}}=-1.862    

    The degrees of freedom are given by:

    df=n-1=11-1=10  

    The p value would be given by:

    p_v =2*P(t_{(10)}<-1.862)=0.0922  

    We see that the p value is higher than the significance level so we don’t have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.

    Step-by-step explanation:

    Information given

    \bar X=15.5 represent the sample mean for the jail time

    s=5.7 represent the sample standard deviation

    n=11 sample size  

    \mu_o =18.7 represent the value that we want to compare

    \alpha=0.05 represent the significance level

    t would represent the statistic

    p_v represent the p value for the test

    System of hypothesis

    We want to check the hypothesis if the true mean for the jail time is equal to 18.7 or no, the system of hypothesis are:

    Null hypothesis:\mu = 18.7  

    Alternative hypothesis:\mu \neq 18.7  

    Since we don’t know the population deviation the statistic is given by:

    t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

    Rreplacing we got:

    t=\frac{15.5-18.7}{\frac{5.7}{\sqrt{11}}}=-1.862    

    The degrees of freedom are given by:

    df=n-1=11-1=10  

    The p value would be given by:

    p_v =2*P(t_{(10)}<-1.862)=0.0922  

    We see that the p value is higher than the significance level so we don’t have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.

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