A supplier of digital memory cards claims that less than 1% of the cards are defective. In a random sample of600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that less than 1% are defective. Use the p-value method.

Answer:

The  decision rule is

Fail to reject the null hypothesis

The conclusion is  

    The is no sufficient evidence to support the claim that the less than 1% of the cards are defective

Step-by-step explanation:

From the question we are told that

     The sample size is  [tex]n = 600[/tex]

      The population proportion of defective samples is p =  0.01

       The sample proportion of defective samples is  [tex]\^ p = 0.03[/tex]

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

The null hypothesis is [tex]H_o : p = 0.01[/tex]

The alternative hypothesis is  [tex]H_a: p < 0.01[/tex]

      Generally the test statistics is mathematically represented as

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

=>              [tex]t = \frac{0.03 - 0.01 }{ \sqrt{ \frac{0.01( 1 - 0.01 )}{600} } }[/tex]

=>              [tex]t = 0.00406[/tex]

From the z table  the area under the normal curve to the left corresponding to  0.00406   is

         [tex]p-value = P(Z < 0.00406 ) = 0.50162[/tex]

comparing the p-valve and the level of significance we see that the

           [tex]p-value > \alpha[/tex]     hence  

The  decision rule is

Fail to reject the null hypothesis

The conclusion is  

    The is no sufficient evidence to support the claim that the less than 1% of the cards are defective

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