Probability involving choosing from objects that are not distinct
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Answer:
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution 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 = 9 - 1 = 8
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.8595
The margin of error is:
M = T*s = 1.8595*2.7 = 5
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 25.3 - 5 = 20.3 milligrams
The upper end of the interval is the sample mean added to M. So it is 25.3 + 5 = 30.3 milligrams.
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.