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alexdok [17]
3 years ago
5

A supplier of digital memory cards claims that less than 1% of the cards are defective. In a random sample of600 memory cards, i

t 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.
Mathematics
1 answer:
Degger [83]3 years ago
3 0

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  n =  600

      The population proportion of defective samples is p =  0.01

       The sample proportion of defective samples is  \^ p = 0.03

       The level of significance is \alpha  = 0.01

       

The null hypothesis is H_o :  p = 0.01

The alternative hypothesis is  H_a: p <  0.01

      Generally the test statistics is mathematically represented as

                  t =  \frac{\^ p - p }{ \sqrt{ \frac{p( 1 - p )}{n} } }      

=>              t =  \frac{0.03  - 0.01 }{ \sqrt{ \frac{0.01( 1 - 0.01 )}{600} } }

=>              t = 0.00406

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

         p-value  = P(Z <  0.00406 ) =  0.50162

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

           p-value > \alpha     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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