Answer:
27 hope it helps you,.................
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:
y=1/3x+2
Step-by-step explanation:
Because it is linear equation in 2 variables
Stephen is likely to qualify since his annual income is less than the median income of Texas.
Random numbers for a simulation might come from
1) a table of random numbers
2) a computer random number generator, such as ones available in many spreadsheet programs
3) digitizing the noise of the Universe. Such random numbers are available on some web sites.