A report from the office of the superintendent claims that the average reading test score of 4th grade students in the school district is 76

Question

A report from the office of the superintendent claims that the average reading test score of 4th grade students in the school district is 76. A group of parents suspects that the real mean is lower than this reported score so they draw a random sample consisting of 37 4th grade student reading exam scores. they find that the sample mean is 74 and the sample standard deviation is 4.5. Find the P-value for the test and state your conclusion at the significance level of 0.05

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Arya 1 week 2021-09-15T23:09:40+00:00 1 Answer 0

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    2021-09-15T23:11:00+00:00

    Answer:

    p_v =P(t_{36}<-2.703)=0.0052    

    If we compare the p value and a significance level given \alpha=0.05 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 scores seems to be lower than 76.    

    Step-by-step explanation:

    Data given and notation    

    \bar X=74 represent the average score for the sample    

    s=4.5 represent the sample standard deviation    

    n=37 sample size    

    \mu_o =76 represent the value that we want to test    

    \alpha=0.05 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 apply a one lower tailed test.  

    What are H0 and Ha for this study?    

    Null hypothesis:  \mu \geq 76  

    Alternative hypothesis :\mu < 76  

    Compute the test statistic  

    The statistic for this case 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=\frac{74-76}{\frac{4.5}{\sqrt{37}}}=-2.703    

    Give the appropriate conclusion for the test  

    First we need to find the degrees of freedom given by:

    df=n-1=37-1=36

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

    p_v =P(t_{36}<-2.703)=0.0052    

    Conclusion    

    If we compare the p value and a significance level given \alpha=0.05 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 scores seems to be lower than 76.    

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