Question

Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 17 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.94 oz and 12.50 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.14 oz. Use the sample data to test the claim that the population of volumes has a standard deviation of less than 0.14 oz. Use a 0.01 significance level.
a. Identify the null and alternative hypotheses.


b. Compute the test statistic. (Round to three decimal places as needed.)


c. Find the p-value. (Round to four decimal places as needed.)


d. State the conclusion.

Answer 1

Answer:

a)Null Hypothesis: $\sigma^2 \geq 0.0196$

Alternative hypothesis: $\sigma^2$

b) $\chi^2 =\frac{17-1}{0.0196} 0.0121 =9.878$

c) $p_v =P(\chi^2$

d)  If we compare the p value we see that is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't say that the true deviation is significantly lower than 0.14

Step-by-step explanation:

Data given

$n=17$ represent the sample size

$\alpha=0.01$ represent the confidence level  

$s^2 =0.11^2 = 0.0121$ represent the sample variance obtained

$\sigma^2_0 =0.14^2 = 0.0196$ represent the value that we want to test

Part a System of hypothesis

On this case we want to check if the population deviation is lower than 0.14, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 0.0196$

Alternative hypothesis: $\sigma^2$

Part b: Calculate the statistic  

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{17-1}{0.0196} 0.0121 =9.878$

Part c: Calculate the p value

In order to calculate the p value we need to find first the degrees of freedom , on this case 17-1=16. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(9.878,16,TRUE)"

Part d: Conclusion

If we compare the p value we see that is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't say that the true deviation is significantly lower than 0.14