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 packages to contain a mean of 8.15 ounces so that the probability of producing a package that contains less than 8 ounces is very small. A sample of 30 packages is selected periodically and weighed, and the packaging process is stopped if there is evidence that the mean packaged amount is statistically different than 8.15 ounces. (They don’t want to under-fill or over-fill.) Suppose that a particular sample of 30 packages had a mean weight of 8.13 ounces, with a sample standard deviation of 0.048 ounces.
Use the appropriate hypothesis test to determine if there is evidence that the population mean is different from 8.15 ounces (use a 0.05 level of significance).
Solution-A:
Ho;mu=8,15
Ha:mu not =8.15
alpha=0.05
Soluiton-b:
test statistic
t=xbar-mu/s/sqrt(n)
t=(8.13-8.15)/(0.048/sqrt(30))
t= -2.282177
t=-2.28
df=n-1=30-1=29
p value==T.DIST.2T(2.282177,29)=0.0300
p<0.05
Reject Ho
Accept Ha
p value gives strong evdience against alternative hypotheis and results are statisticlaly signifcant at 5%
Solution-c:
df=n-1=30-1=29
alpha=0.05
alpha/2=0.05/2=0.025
t crit in excel
=T.INV(0.025,29)
=2.04523
95 %confidence interval for mean is
xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)
8.13-2.04523*0.048/sqrt(30),8.13+2.04523*0.048/sqrt(30)
8.112077,8.147923
we are 95% confidene that the true mean weight lies in between 8.112077 and 8.147923
Solution-d:
there is suffcient statistical evidence at 5% level of significance to conclude that
the population mean is different from 8.15 ounces
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