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 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected perodically, 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 standard deviation of 0.051 ounce.

a. Is there evidence that the population mean amount is different from *.17 ounces? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning.

Answer 1

Answer:

a) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.  

b) Since is a two sided test the p value would be:  

$p_v =2*P(t_{(49)}<-1.525)=0.067$  

Step-by-step explanation:

1) Data given and notation  

$\bar X=8.159$ represent the mean weight for the sample  

$s=0.051$ represent the sample standard deviation

$n=50$ sample size  

$\mu_o =8.17$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 8.17, the system of hypothesis would be:  

Null hypothesis:$\mu = 8.57$  

Alternative hypothesis:$\mu \neq 8.57$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{8.159-8.17}{\frac{0.051}{\sqrt{50}}}=-1.525$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=50-1=49$  

Since is a two sided test the p value would be:  

$p_v =2*P(t_{(49)}<-1.525)=0.067$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.