Question

A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine varies and does not fill each can with exactly 12 ounces. To determine if the filling machine needs adjustment, each day the quality control manager measures the amount of soda per can for a random sample of 50 cans. Experience shows that its filling machines have a known population standard deviation of 0.35 ounces. In today's sample of 50 cans of soda, the sample average amount of soda per can is 12.1 ounces. Construct and interpret a 90% confidence interval estimate for the true population average amount of soda contained in all cans filled today at this bottling plant. Use a 90% confidence level.

Answer 1

Answer:

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$  

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$  

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X=12.1$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=0.35$ represent the population standard deviation  

n=50 represent the sample size  

Solution to the problem  

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$  

Now we have everything in order to replace into formula (1):  

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$  

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$  

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied