Calculation of p-value: Fabric water properties, H 0 : μ ≤ 55 % versus H A : μ &gt; 55 %; n = 15 , x ¯ = 59.81 % , s = 4.94 %I understand that the hypothesized mean is 55 samples mean is 59.81 sample standard deviation is 4.94 sample size is 15. And I use the t-test, I have gotten 3.77106. In this circumstance, how can I get the p-value, and what is different between critical value and p-value?

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Calculation of   p-value: Fabric water properties,       H    0    :  μ  ≤  55  % versus       H    A    :  μ  &amp;gt;  55  %;  n  =  15  ,      x    ¯    =  59.81  %  ,  s  =  4.94  %I understand that the hypothesized mean is   55 samples mean is   59.81 sample standard deviation is   4.94 sample size is   15. And I use the   t-test, I have gotten   3.77106. In this circumstance, how can I get the   p-value, and what is different between critical value and   p-value?

svirajueh · Accepted Answer

The PDF of a Student   t distribution with   ν  =  n  −  1  =  14 degrees of freedom is      f    T    (  t  )  =      2473099629    16              (              1                              t            2                    +          14                    )              15          /      2        .The desired probability is therefore  p  =  Pr  [  T  &amp;gt;  3.77106  ]  =      ∫          t      =      3.77106        ∞        f    T    (  t  )    d  t  ,which you can either numerically integrate, or it is integrable in closed form as                    1        −                  F          T                (                  t          0                )                            =                  ∫                      t            =                          t              0                                ∞                          f          T                (        t        )                d        t                                          =                  1          2                −                              16                          t              0                          13                        +            1456                          t              0                          11                        +            56056                          t              0              9                        +            1177176                          t              0              7                        +            14420406                          t              0              5                        +            100942842                          t              0              3                        +            353299947                          t              0                                            32                                          (                                  t                  0                  2                                +                14                )                                            13                              /                2                                                        which gives  Pr  [  T  &amp;gt;  3.77106  ]  =  1  −      F    T    (  3.77106  )  ≈  0.00103279.Usually, however, we use computer programs or statistical tables to compute these   p-values, since the larger the sample size, the more impractical it becomes to evaluate the integral in closed form.As for your question about the difference between   p-value and critical value, the   p-value is in your case the probability that the statistic   T is at least as large as the value of the statistic that was observed. That is to say, you are trying to quantify how &quot;unusual&quot; it is to observe the result you computed, assuming the null hypothesis is true. If it is highly unusual (low probability), this means the null hypothesis is unlikely to be true.It&#039;s like saying, if I assume a coin is fair and I toss it   100 times and get   90 heads and   10 tails, is it now likely that my assumption is correct? The way you formalize your conclusion is that if the coin were fair, the probability of obtaining such an extreme result is very small. The   p-value is this probability. But if you had assumed that the coin has a probability of   0.9 of landing heads up,   90 heads and   10 tails would not be unusual at all.The critical value is another way of making a decision about a hypothesis test. Instead of a probability, a critical value is a quantile: in your case, it is some value       t      crit   such that  Pr  [  T  &amp;gt;      t      crit    ]  =  0.05  ,which is the       95          t      h       percentile of the Student t distribution with   14 degrees of freedom, which using a computer, is about       t      crit    ≈  1.76131. That means if your test statistic is larger than this value, you would reject       H    0  . It also means for any   T  &amp;gt;  1.76131, the   p-value is less than   0.05.

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