sample of 60 account balances of a credit company showed an average balance of $1,165 and a standard deviation of $125. You want to determin

Question

sample of 60 account balances of a credit company showed an average balance of $1,165 and a standard deviation of $125. You want to determine if the mean of all account balances is significantly greater than $1,150. Assume the population of account balances is normally distributed. Compute the p-value for this test.

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Amara 1 day 2021-10-13T02:20:55+00:00 1 Answer 0

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    2021-10-13T02:22:46+00:00

    Answer:

    z=0.930

    p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176

    Step-by-step explanation:

    1) Data given and notation    

    \bar X=1165 represent the mean for the account balances of a credit company

    s=125 represent the population standard deviation for the sample    

    n=60 sample size    

    \mu_o =1150 represent the value that we want to test  

    \alpha represent the significance level for the hypothesis test.  

    z would represent the statistic (variable of interest)    

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

    2) State the null and alternative hypotheses.    

    We need to conduct a hypothesis in order to determine if the mean for account balances of a credit company is greater than 1150, the system of hypothesis would be:    

    Null hypothesis:\mu \leq 1150    

    Alternative hypothesis:\mu > 1150    

    We don’t know the population deviation, but the problem says the the distribution for the random variable is normal, so for this case we can use the z test to compare the actual mean to the reference value, and the statistic is given by:    

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

    z-test: “Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.

    3) Calculate the statistic    

    We can replace in formula (1) the info given like this:    

    z=\frac{1165-1150}{\frac{125}{\sqrt{60}}}=0.930    

    4) Calculate the P-value    

    Since is a one-side upper test the p value would be:    

    p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176

    In Excel we can use the following formula to find the p value “=1-NORM.DIST(0.93,0,1,TRUE)”  

    5) Conclusion    

    If we compare the p value with a significance level for example \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the balances of a credit company are significantly higher than $1150 at 0.05 of signficance.    

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