According to the US Bureau of labor statistics, 7% of US female workers between 16 and 24 years old are paid at the minimum wage or less. A state politician wants to verify this statement for his state. He uses a sample of 500 female workers and finds 42 are paid at the minimum wage or less. Use a 5% significance level to test to test whether that state differs from the nation.

State clearly the null and the alternative hypothesis, the test statistic, the decision rule and the conclusion.

Respuesta :

Answer:

The Null Hypothesis is  [tex]H_o:k_o = 0.07[/tex]

The alternative hypothesis is  [tex]H_a :k_o \ne 0.07[/tex]

Decision rule

    If the test staistics is  greater than the critical value of significance level then [tex]H_o[/tex] is accepted else [tex]H_o[/tex]  is rejected

With the above in mind

       The critical value of the significance level which is obtained from the table is

     [tex]t_{0.05} = 1.645[/tex]

Now since the critical value of significance level is greater than the test  staistics then the null hypothesis will be rejected

Conclusion

The information is not enough to back the claim that state differs from the nation

Step-by-step explanation:

From the question we are told that

The percentage of US female workers paid at the minimum wage or less is  [tex]k_o =[/tex] 7% = 0.07

  The sample size is  [tex]n = 500[/tex]

    The number paid minimum wage or less is  x = 42

    The significance level is [tex]\alpha =[/tex]5%  = 0.05

Now the probability of getting a US female workers paid at the minimum wage or less is mathematically represented as

         [tex]\= k = \frac{x}{n}[/tex]

substituting value

        [tex]\= k = \frac{42}{500}[/tex]

        [tex]\= k = 0.084[/tex]

The Null Hypothesis is  [tex]H_o:k_o = 0.07[/tex]

The alternative hypothesis is  [tex]H_a :k_o \ne 0.07[/tex]

Generally the test statistics is mathematically evaluated as

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

substituting value

      [tex]z = \frac{0.084 - 0.07}{\sqrt{\frac{0.07 (1-0.07)}{500} } }[/tex]

     [tex]z = 1.23[/tex]

Now the Decision rule is stated as  

        If the test staistics is  greater than the critical value of significance level then [tex]H_o[/tex] is accepted else [tex]H_o[/tex]  is rejected

With the above in mind

       The critical value of the significance level which is obtained from the table is

     [tex]t_{0.05} = 1.645[/tex]

Now since the critical value of significance level is greater than the test  staistics then the null hypothesis will be rejected

   So the conclusion will be

   The information is not enough to back the claim that state differs from the nation

 

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