Complete Question
A sample of 1200 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that more than 50% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.10 level to support the company's claim
Answer:
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to conclude that the company's promotional literature claimed that more than 50% do not fail in the first 1000 hours of their use is correct
Step-by-step explanation:
From the question we told that
The sample size is n = 1200
The significance level is [tex]\alpha = 0.100[/tex]
The sample proportion is [tex]\^ p = 0.53[/tex]
The null hypothesis [tex]H_o : p = 0.50[/tex]
The alternative hypothesis is [tex]Ha : p > 0.50[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{\^ p - p }{ \sqrt{\frac{p(1 - p )}{n} } }[/tex]
=> [tex]z = \frac{ 0.53 - 0.5 }{ \sqrt{\frac{0.50 (1 - 0.50)}{1200} } }[/tex]
=> [tex]z = 2.078[/tex]
From the z table the area under the normal curve to the right corresponding to 2.078 is
[tex]P(Z > 2.078 ) = 0.018855[/tex]
So
[tex]p-value = 0.018855[/tex]
Generally the value obtained we see that [tex]p-value < \alpha[/tex] , Hence
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to conclude that the company's promotional literature claimed that more than 50% do not fail in the first 1000 hours of their use is correct
