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Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance alpha using the given sample statistics. Claim: p is not equal to 0.22; Alpha = 0.01; Sample​ statistics: ​p = 0.15, n = 180Can the normal sampling distribution be​ used? A. ​No, because np is less than 5. B. ​No, because nq is less than 5. C. ​Yes, because both np and nq are greater than or equal to 5. D. ​Yes, because pq is greater than alpha = 0.01. State the null and alternative hypotheses. Determine the critical value(s).Find the z-test statistic.

Answer :

okpalawalter8

Answer:

A

   The correct option is  C

B

  The  value of  z-test is  [tex]z = -2.267[/tex]

Step-by-step explanation:

From the question we are told that  

    The null hypothesis  [tex]H_o : p = 0.22[/tex]

    The alternative hypothesis  [tex]H_a : p \ne 0.22[/tex]

    The  level of significance is  [tex]\alpha = 0.01[/tex]

    The sample proportion is  [tex]\^ p = 0.15[/tex]

     The sample size is  n = 180

Generally from central limit theorem

if   np  and  nq  are  >  5  then  normal sampling distribution can be used

So  

     np =  180  * 0.22 =  39.6  >  5

and  

     nq =  180 * (1 -0.22) =  140.4 > 5

So normal sampling distribution can be used

Generally the z-test is mathematically represented as

      [tex]z = \frac{\^ p - p }{ \sqrt{\frac{ p(1- p ) }{n } } }[/tex]

=>   [tex]z = \frac{ 0.15 - 0.22 }{ \sqrt{\frac{ 0.2(1- 0.22 ) }{ 180 } } }[/tex]

=>   [tex]z = -2.267[/tex]

   

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