Question

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).

- State the null and alternative hypothesis.
- Determine the p-value and interpret its meaning
- Construct a 95% confidence interval of the mean population weight and interpret its meaning.
- Based on the information in (a) and (b), what is your decision as process manager?

Answer #1

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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