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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.
Answer:
The value is
Step-by-step explanation:
From the question we are told that
The population proportion is
The sample size is n = 563
Generally the population mean of the sampling distribution is mathematically represented as
Generally the standard deviation of the sampling distribution is mathematically evaluated as
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Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as
Here is the sample proportion of persons with a college degree.
So
Here
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From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is
and
So
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