The amounts of nicotine in a certain brand of cigarette are normally distributed

Question

with a mean of 0.896 g and a standard deviation of 0.329 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 43 cigarettes with a mean nicotine amount of 0.831 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 43 cigarettes with a mean of 0.831 g or less. P(M < 0.831 g) =

Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Assume that any probability below 5% is sufficient to conclude that the new product really have lower nicotine. Based on the result above, is it valid to claim that the amount of nicotine is lower?

answer is 0.0976 or 0.0975

is there anyway to calculate this by hand,calculator or excel?

Answer 1

Hi Cynthya,

The by-hand method will give you a slightly different answer because z-tables are not exact. To reproduce the answer you specify, we can use Excel. We will need the sample mean, population mean, and standard error. Sample mean is 0.831, population mean is 0.896, and we can compute standard error with the following formula:

SE=sigma/sqrt(n)

sigma=standard deviation=0.329

n=sample size=43

SE=0.329/sqrt(43)

SE=0.0502

Now, we have all values we need for Excel:

xbar=0.831

mu=0.896

SE=0.0502

Go to Excel and type the following formula into any cell:

=NORM.DIST(0.831,0.896,0.0502,TRUE)

This will give you approximately 0.0976. I hope this helps.