For H 0 : π = 0.20, for instance, the score test statistic is Z ( s ) = − 2.50, which has two-sided P-value 0.012 &lt; 0.05, so 0.20 does not fall in the interval.It is understood that the score test statistic is − 2.5 using n = 25 and π ^ = 0 (in the previous description), but the part where the P-value is 0.012 is not understood.

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For       H    0    :  π  =  0.20, for instance, the score test statistic is   Z  (  s  )  =  −  2.50, which has two-sided P-value   0.012  &amp;lt;  0.05, so   0.20 does not fall in the interval.It is understood that the score test statistic is   −  2.5 using   n  =  25 and       π    ^    =  0 (in the previous description), but the part where the P-value is   0.012 is not understood.

charringpq49u · Accepted Answer

The   2-sided P-value is double the probability below   −  2.5 under the standard normal density curve; that is,  P  (  |  Z  |  &amp;gt;  2.5  )  =  P  (  Z  &amp;lt;  −  2.5  )  +  P  (  Z  &amp;gt;  2.5  )  =  2  P  (  Z  &amp;lt;  −  2.5  )    =  2  (  0.0062  )  =  0.0124  &amp;lt;  0.05  =  5  %  ,so the null hypothesis is rejected at the   5  % level.In   R, where   p  n  o  r  m is a normal CDF:  p  n  o  r  m  (  −  2.50  )  [  1  ]     0.006209665Or use printed standard normal CDF tables (usually limited to   4-place accuracy).

For H 0 </msub> : π = 0.20 , for instance, the score test sta

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