Question

The maker of potato chips uses an automated packaging machine to pack its 20-ounce bag of chips. At the end of every shift 18 bags are selected at random and tested to see if the equipment needs to be readjusted. After one shift, a sample of 18 bags yielded the following data. mean = 20.45 s = .80 n = 18. If we were to conduct a test to see if the sample estimate is different from the company’s expected weight of 20 oz. at an alpha level = .05, what is the conclusion for this test?

Answer 1

Answer with explanation:

Let $\mu$ be the population mean .

By considering the given information in the question , we have

$\text{Null hypothesis }H_0: \mu=20\\\\\text{Alternative hypothesis } H_a: \mu\neq20$

Here, $H_a$ is two-tailed , so the test is a two-tailed test.

Also, population standard deviation is unknown , so we perform  a two-tailed t-test.

For sample size : n= 18

Sample mean : $\overline{x}=20.45$

Sample standard deviation : s=0.80

Test statistic :

$t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$t=\dfrac{20.45-20}{\dfrac{0.80}{\sqrt{18}}}$

$t=\dfrac{0.45}{\dfrac{0.80}{4.2426}}$

$t=\dfrac{0.45}{0.1885636}$

$t\approx 2.39$

Two-tailed Critical values for $\alpha=0.05$ and degree of freedom df=n-1=17  

$t_{(\alpha/2,\ df)}=t_{(0.025,\ 17)}=\pm2.11$

Decision : Since the calculated t-value(2.39) does not lie between the critical values -2.11 and 2.11.

So we reject the null hypothesis .

Conclusion : We have enough evidence at 0.05 significance level to support the claim that the sample estimate is different from the company’s expected weight of 20 oz.