Given a test statistic of t=2.315 of a left tailed test with n=8

Question

Given a test statistic of t=2.315 of a left tailed test with n=8

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Amaya 1 week 2021-11-18T03:41:36+00:00 1 Answer 0 views 0

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    2021-11-18T03:43:03+00:00

    Answer:

    Null hypothesis:  \mu =110

    Alternative hypothesis:\mu \neq 110

    The sample size on this case is n=8, then the degrees of freedom are given by:

     df = n-1= 8-1=7

    The statistic is given by:

     t= \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

    For this case the value of the statistic is given  t = 2.315

    Since we are using a bilateral test the p value would be given by:

     p_v = 2*P(t_{7}>2.315) =0.054

    And we can use the following excel code to find it:

    “=2*(1-T.DIST(2.315;7;TRUE))”

    Since the p value is higher than the significance level given we FAIL to reject the null hypothesis. And the best conclusion would be:

    0.05<P-value <0.10, fail to reject the null hypothesis

    Step-by-step explanation:

    Assuming this complete question :”Given a test statistic of t=2.315 of a left-tailed test with n=8, use a 0.05 significance level to test a claim that the mean of a given population is equal to 110.

    Find the range of values for the P-value and state the initial conclusion 1 point) 0.05<P-value <0.10; reject the null hypothesis

    0.05<P-value <0.10, fail to reject the null hypothesis

    0.025 < P-value <0.05; reject the null hypothesis

    0.025< P-value<0.05; fail to reject the null hypothesis

    For this case they want to test if the population mean is 110 or no, the systemof hypothesis are:

    Null hypothesis:  \mu =110

    Alternative hypothesis:\mu \neq 110

    The sample size on this case is n=8, then the degrees of freedom are given by:

     df = n-1= 8-1=7

    The statistic is given by:

     t= \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

    For this case the value of the statistic is given  t = 2.315

    Since we are using a bilateral test the p value would be given by:

     p_v = 2*P(t_{7}>2.315) =0.054

    And we can use the following excel code to find it:

    “=2*(1-T.DIST(2.315;7;TRUE))”

    Since the p value is higher than the significance level given we FAIL to reject the null hypothesis. And the best conclusion would be:

    0.05<P-value <0.10, fail to reject the null hypothesis

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