A sample of 1600 computer chips revealed that 50% of the chips fail in the first 1000 hours of their use. The company's promotional literature claimed that under 53% fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim? State the null and alternative hypotheses for the above scenario.
Solution :
Given that,
= 0.53
1 - = 0.47
n = 1600
Level of significance = = 0.02
Point estimate = sample proportion = = 0.50
This a left (One) tailed test.
The null and alternative hypothesis is,
Ho: p = 0.53
Ha: p < 0.53
Test statistics
z = ( - ) / *(1-) / n
= ( 0.50 - 0.53) / (0.53*0.47) /1600
= -2.404
P-value = P(Z < z)
= P(Z < -2.404 )
= 0.0081
The p-value is p = 0.0081, and since p = 0.0081 < 0.02, it is concluded that the null hypothesis is rejected.
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