Another one that I need help....
At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 520 ml. The filling process follows a normal distribution with a known process standard deviation of 4 ml. (a) Which sampling distribution would you use if random samples of 10 bottles are to be weighed? Why? (b) Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance. (c) If a sample of 16 bottles shows a mean fill of 515 ml, does this contradict the hypothesis that the true mean is 520 ml?
Okay, never mind I think I got this one....
To answer these questions, we need to understand the concept of sampling distribution, hypothesis testing, and the t-distribution. Let's break down each question step by step:
(a) Which sampling distribution would you use if random samples of 10 bottles are to be weighed? Why?
In this case, since we are dealing with a sample size of 10 bottles and we know the population standard deviation (4 ml), we can use the normal distribution for the sampling distribution. When the sample size is relatively large (n > 30) or the population standard deviation is known, the sampling distribution tends to be approximately normally distributed.
(b) Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance.
The hypotheses can be set up as follows:
- Null hypothesis (H0): The true mean fill of the bottles is 520 ml.
- Alternative hypothesis (Ha): The true mean fill of the bottles is not equal to 520 ml.
The two-tailed decision rule for the 5 percent level of significance means that we will reject the null hypothesis if the sample mean is either significantly greater than or significantly less than 520 ml. The critical region will be split into two equal tails with each tail having an alpha level of 2.5 percent.
(c) If a sample of 16 bottles shows a mean fill of 515 ml, does this contradict the hypothesis that the true mean is 520 ml?
To determine whether the sample of 16 bottles contradicts the hypothesis, we need to calculate the t-statistic and compare it to the critical value using the t-distribution.
First, we calculate the standard error of the mean using the formula:
Standard Error = Population Standard Deviation / √(Sample Size)
Standard Error = 4 ml / √(16) = 1 ml
Next, we calculate the t-statistic using the formula:
t = (Sample Mean - Population Mean) / Standard Error
t = (515 ml - 520 ml) / 1 ml = -5 / 1 = -5
Finally, we compare the absolute value of the t-statistic (-5) with the critical value from the t-distribution table. Since the sample size is 16 and we are using a two-tailed test at a 5 percent level of significance (α = 0.05), the critical value is approximately 2.131.
Since the absolute value of the t-statistic (-5) is greater than the critical value (2.131), we can reject the null hypothesis. This means that the sample of 16 bottles contradicts the hypothesis that the true mean is 520 ml.
In conclusion, based on the provided information, the sample of 16 bottles with a mean fill of 515 ml contradicts the hypothesis that the true mean is 520 ml.