A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects and defects occur
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Answer:
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 27 - 1 = 26
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.7787.
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.
The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).