The county health department has concerns about the chlorine level of 0.4% mg/mL at a local water park increasing to an unsafe level. The wa

Question

The county health department has concerns about the chlorine level of 0.4% mg/mL at a local water park increasing to an unsafe level. The water department tests the hypothesis that the local water park’s chlorine proportions have remained the same, and find a P-value of 0.005.
Provide an appropriate conclusion.

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Iris 3 weeks 2021-09-25T21:04:46+00:00 1 Answer 0

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    2021-09-25T21:06:15+00:00

    Answer:

    p_v =P(t_{(n-1)}>t_{calc})=0.005  

    If we compare the p value and the significance level given assumed \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 true mean is higher than 0.4 % mg/mL at any significance level higher than 0.5%

    Step-by-step explanation:

    Data given and notation  

    \bar X represent the sample mean

    s represent the sample standard deviation

    n sample size  

    \mu_o =0.4 represent the value that we want to test

    \alpha represent the significance level for the hypothesis test.  

    t would represent the statistic (variable of interest)  

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

    State the null and alternative hypotheses.  

    We need to conduct a hypothesis in order to check if the mean i higher than 0.4, the system of hypothesis would be:  

    Null hypothesis:\mu \leq 0.4  

    Alternative hypothesis:\mu > 60.4  

    We assume that we don’t know th population deviation to we need to apply a t test, and the statistic is given by:

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

    t-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”.  

    Calculate the statistic

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

     t_{calc}

    P-value

    The first step is calculate the degrees of freedom, on this case:  

    df=n-1  

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

    p_v =P(t_{(n-1)}>t_{calc})=0.005  

    Conclusion  

    If we compare the p value and the significance level given assumed \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 true mean is higher than 0.4 % mg/mL at any significance level higher than 0.5%

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