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grandpianograce
03.03.2020 •
Mathematics
A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tests them to make sure that they do not have any dead pixels. Over the course of a month, she tested 300 monitors and found 24 monitors with dead pixels. The technician conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of monitors with dead pixels is greater than 5%.
a. H0:p=0.05; Ha:p>0.05, which is a right-tailed test.
b. Use Excel to test whether the true proportion of monitors with dead pixels is greater than 5%. Identify the test statistic, z, and p-value, rounding to three decimal places.
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Ответ:
There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 300
p = 5% = 0.05
Alpha, α = 0.05
Number of dead pixels , x = 24
First, we design the null and the alternate hypothesis
This is a one-tailed(right) test.
Formula:
Putting the values, we get,
Now, we calculate the p-value from excel.
P-value = 0.00856
Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.
Ответ:
No, the work is not correct. The distributive property was not used properly to expand the expression. Negative 6 should be multiplied by both terms in the parentheses. The second product should be -6 times -2/13 to get positive 12/13.
Step-by-step explanation: THAT IS THE SAMPLE RESPONSE!!! :)