Question

A salesman for a new manufacturer of cellular phones claims not only that they cost the retailer less but also that the percentage of defective cellular phones found among his products, ( p1), will be no higher than the percentage of defectives found in a competitor's line, ( p2 ). To test this statement, the retailer took a random sample of 135 of the salesman's cellular phones and 175 of the competitor's cellular phones. The retailer found that 14 of the salesman's cellular phones and 8 of the competitor's cellular phones were defective. Does the retailer have enough evidence to reject the salesman's claim? Use a significance level of α = 0.01 for the test. 1 of 6: State the null and alternative hypotheses for the test Step 2 of 6: Find the values of the two sample proportions, pˆ1 and pˆ2. Round your answers to three decimal places. Step 3 of 6: Compute the weighted estimate of p, ‾‾p. Round your answer to three decimal places Step 4 of 6: Compute the value of the test statistic. Round your answer to two decimal places. Step 5 of 6: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places. Step 6 of 6: Make the decisi on for the hypothesis test. Reject or Fail to Reject Null Hypothesis.

Answer 1

Here, we have given that,

Step 1 of 6

Claim: To check whether the percentage of defective cellular phones found among cellular phones, ( p1), will be no higher than the percentage of defectives found in a competitor's line, ( p2 ).

The null and alternative hypotheses are as follows:

v/s

Where p1 = The percentage of defective cellular phones found cellular phones

p2= the percentage of defectives found in a competitor's line.

This is the left- one-tailed test.

Here, we are using the two-sample proportion test.

Step 2 of 6

n1= number of salesman's cellular phones=135

x1= Number of salesman's cellular phones defective = 14

=Sample proportion of salesman's cellular phones defective=

n2= number of competitor's cellular phones= 175

x2=number of competitor's cellular phones defective = 8

=Sample proportion of competitor's cellular phones defective

Step 3 of 6

The weighted estimate of p is

= = =0.075

Step 4 of 6

Now, we can find the test statistics

Z-statistics=

                 =

       =1.89

The test statistics is 1.89

Step 5 of 6

Now, we can find the p-value

P-value = P( Z< z) as this is left one-tailed test

              =P (Z < 1.89)

              =0.9706 Using standard normal z table see the value corresponding to the z=1.89

we get the p-value is 0.9706

Decision Rule:

=level of significance =0.05

Reject Ho null hypothesis if the P-value < 0.05 otherwise we fail to reject Ho Null hypothesis.

Step 6 of 6

Conclusion:

Here P-value (0.9706) > 0.05

we fail to reject the Null hypothesis Ho

we conclude that there is not sufficient evidence to support the claim the percentage of defective cellular phones found among cellular phones, ( p1), will be no higher than the percentage of defectives found in a competitor's line, ( p2 ).