A average amount of money on deposit in a savings account is $7500. Suppose a random sample of 49 accounts shows the average in their bank t

Question

A average amount of money on deposit in a savings account is $7500. Suppose a random sample of 49 accounts shows the average in their bank to be $6850 with a population s.d. $1200. Construct a hypothesis test to determine whether the average amount on deposit per account is different from $7500. Use a 1% level of significance. Use the p value approach.

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Aaliyah 1 week 2021-10-12T07:22:38+00:00 1 Answer 0

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    2021-10-12T07:23:45+00:00

    Answer:  

    z=-3.79

    p_v =2*P(z<-3.79)=0.000151  

    Step-by-step explanation:  

    1) Data given and notation  

    \bar X=6850 represent the mean average amount of money on deposit in savings account for the sample  

    \sigma=1200 represent the population standard deviation for the sample  

    n=49 sample size  

    \mu_o =7500 represent the value that we want to test

    \alpha=0.01 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 check if the mean for the amount of money on deposit in savings account differs from 7500, the system of hypothesis would be:  

    Null hypothesis:\mu =7500  

    Alternative hypothesis:\mu \neq 7500  

    Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

    z=\frac{\bar X-\mu_o}{\frac{\sigma}{\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{6850-7500}{\frac{1200}{\sqrt{49}}}=-3.79  

    4)P-value  

    Since is a two-sided test the p value would be:  

    p_v =2*P(z<-3.79)=0.000151  

    5) Conclusion  

    If we compare the p value and the significance level given \alpha=0.01 we see that p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average amount of money deposit in savings account its significantly different from 7500 tons at 1% of signficance.  

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