A machine fills 12 ounce bottles with soda. For the machine to function properly, the standard deviation of the sample must be less than or equal to 0.02 ounce. A sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At alpha +0.05, can you reject the claim that the machine is functioning properly? data is 12.06 12.02 11.98 12.05 12.04 12.09 11.99 12.04
The standard deviation of the sample is 0.03, which is greater than 0.02. Therefore, you can reject the claim that the machine is functioning properly at alpha +0.05.
To determine whether the claim can be rejected, we need to conduct a hypothesis test. Let's go step-by-step:
Step 1: State the hypotheses.
The null hypothesis (H0): The machine is functioning properly (standard deviation is less than or equal to 0.02 ounce).
The alternative hypothesis (H1): The machine is not functioning properly (standard deviation is greater than 0.02 ounce).
Step 2: Formulate the test statistic.
Since the sample standard deviation is unknown, we need to use a t-test statistic. The test statistic for this hypothesis test is given by:
t = (s - σ) / (s / √n)
where:
s = sample standard deviation = 0.0338 (calculated from the given data)
σ = hypothesized standard deviation = 0.02 (given)
n = sample size = 8 (given)
Let's calculate the test statistic:
t = (0.0338 - 0.02) / (0.0338 / √8) = 1.57
Step 3: Determine the critical value.
Since the significance level (α) is given as 0.05, and this is a one-tailed test with the alternative hypothesis stating that the standard deviation is greater than 0.02, we need to find the critical value associated with a 0.05 significance level and degrees of freedom (df) equal to (n - 1).
Consulting the t-distribution table or using statistical software, we find the critical value to be approximately 1.860.
Step 4: Make a decision.
If the calculated test statistic (1.57) is greater than the critical value (1.860), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since 1.57 < 1.860, we fail to reject the null hypothesis.
Step 5: State the conclusion.
Based on the given sample data and a significance level of 0.05, there is not enough evidence to reject the claim that the machine is functioning properly (standard deviation less than or equal to 0.02 ounce).
To determine whether we can reject the claim that the machine is functioning properly, we need to conduct a hypothesis test. In this case, we will be performing a one-sample t-test.
The null hypothesis, denoted as H0, assumes that the machine is functioning properly, meaning the mean of the sample is equal to 12 ounces. The alternative hypothesis, denoted as Ha, assumes that the machine is not functioning properly, meaning the mean of the sample is different from 12 ounces.
H0: μ = 12 (sample mean equals 12 ounces)
Ha: μ ≠ 12 (sample mean is not equal to 12 ounces)
To perform the test, we will calculate the t-statistic, which measures how much the sample mean differs from the assumed population mean. Here are the steps to calculate the t-statistic:
1. Calculate the sample mean (x̄) of the ounces of soda in the 8 bottles.
x̄ = (12.06 + 12.02 + 11.98 + 12.05 + 12.04 + 12.09 + 11.99 + 12.04) / 8 = 12.035
2. Calculate the sample standard deviation (s) of the ounces of soda in the 8 bottles.
s = √[(Σ(xi - x̄)²) / (n - 1)]
s = √[((12.06 - 12.035)² + (12.02 - 12.035)² + (11.98 - 12.035)² + (12.05 - 12.035)² + (12.04 - 12.035)² + (12.09 - 12.035)² + (11.99 - 12.035)² + (12.04 - 12.035)²) / (8 - 1)]
s ≈ √[0.00196 / 7] ≈ 0.020
3. Calculate the standard error (SE).
SE = s / √n
SE = 0.020 / √8 ≈ 0.007
4. Calculate the t-statistic.
t = (x̄ - μ) / SE
t = (12.035 - 12) / 0.007 ≈ 5
Now that we have calculated the t-statistic, we can compare it with the critical t-value at alpha +0.05 (or 0.025 on each tail) with 7 degrees of freedom.
Using a t-table or calculator, we find that the critical t-value is approximately 2.365.
Since the calculated t-statistic (5) is greater than the critical t-value (2.365) in absolute value, we can reject the null hypothesis.
In conclusion, at alpha +0.05, we can reject the claim that the machine is functioning properly based on the provided data.