Question 1169145: Workers at a certain soda drink factory collected data on the volumes (in ounces) of a simple random sample of 21 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.12 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 11.87 oz and 12.55 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.17 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.17 oz. Use a 0.05 significance level. Complete parts (a) through (d)
h0: o=0.17
h1: o< 0.17
need to find x to the second power
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Absolutely, let's break down this hypothesis test step-by-step.
**a) State the Hypotheses**
* $H_0: \sigma = 0.17$ (The population standard deviation is 0.17 oz)
* $H_1: \sigma < 0.17$ (The population standard deviation is less than 0.17 oz)
**b) Calculate the Test Statistic (Chi-Square)**
We'll use the chi-square test statistic for standard deviation, given by:
$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$
Where:
* $n$ is the sample size (21)
* $s$ is the sample standard deviation (0.12 oz)
* $\sigma_0$ is the hypothesized population standard deviation (0.17 oz)
Let's plug in the values:
$\chi^2 = \frac{(21-1)(0.12)^2}{(0.17)^2} = \frac{(20)(0.0144)}{0.0289} = \frac{0.288}{0.0289} \approx 9.9654$
So, $\chi^2 \approx 9.9654$
**c) Determine the Critical Value**
* We have a left-tailed test (because $H_1: \sigma < 0.17$).
* The significance level is $\alpha = 0.05$.
* The degrees of freedom are $df = n - 1 = 21 - 1 = 20$.
We need to find the critical chi-square value $\chi^2_{\text{critical}}$ such that the area to the left of it is 0.05.
Using a chi-square distribution table or calculator, we find that:
$\chi^2_{\text{critical}} \approx 10.851$
**d) Make a Decision and Draw a Conclusion**
* **Decision Rule:** If $\chi^2 < \chi^2_{\text{critical}}$, reject $H_0$.
* We have $\chi^2 \approx 9.9654$ and $\chi^2_{\text{critical}} \approx 10.851$.
* Since 9.9654 < 10.851, we reject $H_0$.
**Conclusion:**
There is sufficient evidence at the 0.05 significance level to support the claim that the population of volumes has a standard deviation less than 0.17 oz.
**Therefore:**
* $\chi^2 \approx 9.9654$
|
|
|
