SOLUTION: The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.883 g and a standard deviation of 0.289 g. The company that produces these cigaret

Algebra ->  Probability-and-statistics -> SOLUTION: The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.883 g and a standard deviation of 0.289 g. The company that produces these cigaret      Log On


   

Question 1202805: The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.883 g and a standard deviation of 0.289 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 33 cigarettes with a mean nicotine amount of 0.777 g.
Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 33 cigarettes with a mean of 0.777 g or less.
P(M < 0.777 g) =
Enter your answer as a number accurate to 4 decimal places.
Based on the result above, is it valid to claim that the amount of nicotine is lower?
No. The probability of obtaining this data is high enough to have been a chance occurrence.
Yes. The probability of this data is unlikely to have occurred by chance alone.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population mean = .883
population standard deviation = .289
sample size = 33
sample mean = .777
you want to find the probability of getting an amount of nicotine less than .777 from this sample.
z-score formula is z = (x-m) / s
z is the z-score
x is the sample mean
m is the population mean
s is the standard error.
standard error = population standard deviation divied by square root of sample size = .289 / sqrt(33) = .0503.
z-score formula becomes z = (.777 - .883) / .0503 = -2.1074.
the probability of getting a z-score less than that is equal to .0275.
that's a pretty low probability.
if you were tyesting at .05 one tail level of significance, the results would be considered significant and you would conclude that the probability of getting a z-score this low is leess likely to be due to random variations in the mean of a sample of this size and more likely that the actual meana really is lower.
if you were testing at .01 one tail level of significance, the resuls would be considered not significant and you would conclude that the proability of getting a z-sore this low is more likely to be due to random variations in the mean of a sample of this size and so you would considered the assumed population mean to be valid.
i ran this through a one sample z-test calculator and it came up with the same conclusion.
here are the results.





the first is significant at .05 level of significance.
the second is not significance at .01 level of significance.