question archive A manufacturer of chocolate candies uses machines to package candies as they move along a filling line
A manufacturer of chocolate candies uses machines to package candies as they move along a filling line
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A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the pack-ages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages, the mean amount dispensed is 8.159 ounces, with a sample stan-dard deviation of 0.051 ounce.
a. Is there evidence that the population mean amount is different from 8.17 ounces? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning
Solution needs to be done in excel. Please assist with the formulas
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H0: u= 8.17
H1: u≠8.17
test: this is a case of z test:
By manual , z= (sample mean - population mean)/standard deviation /(sqrt (n):
z= (8.159-8.17)/0.051/sqrt(50)= -1.53
The z statistic attained is then subject to the normal distribution table where the p-value is determined in this case 0.1272 ( two tailed)
This is however automated by excel using =norm.dist(x,mean,standard deviation,cumulative) function which automatically returns the p -value as shown.
=NORM.DIST(8.159,8.17,0.051/SQRT(50),1)
Excel returns a p-value of 0.063613
since the test is two tailed test, the p-value is multiplied by 2:
As such the formula is modified as :
=NORM.DIST(8.159,8.17,0.051/SQRT(50),1)
= 0.127226
Decision rule: reject null if p < .05
Since p > .05 we fail to reject null
conclusion
There is no sufficient evidence to suggest that the mean is different from 8.17 ounces .
