Exercise :
A machine is set up such that the average content of juice per bottle equals μ . A sample of 36 bottles yeilds an average content of 48.5cl. Test the hypothesis that the average content per bottle is 50cl at the significance level.
Assume that the population standard deviation σ = 5cl
To test the hypothesis that the average content per bottle is 50cl at a given significance level, we can perform a hypothesis test using the information given.
Here's how to approach it:
1. State the null hypothesis (H0) and the alternative hypothesis (H1):
- Null hypothesis (H0): The average content per bottle is μ = 50cl.
- Alternative hypothesis (H1): The average content per bottle is not equal to 50cl (μ ≠ 50cl).
2. Determine the significance level (α), which is typically provided in the question. Let's assume α = 0.05 (5%).
3. Calculate the test statistic:
- To calculate the test statistic, we use the formula:
t = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean (in this case, 50cl), σ is the population standard deviation (5cl), and n is the sample size (36 bottles).
Plugging in the values from the question, we have:
t = (48.5 - 50) / (5 / √36)
t = -1.5 / (5 / 6)
t = -1.5 * (6 / 5)
t = -1.8
4. Determine the critical values for the test statistic:
- Since the alternative hypothesis is two-tailed (μ ≠ 50cl), we need to divide the significance level (α) by 2 to get the critical values.
- Look up the critical values for the t-distribution with degrees of freedom (df) equal to (n - 1) = (36 - 1) = 35. For α/2 = 0.05/2 = 0.025 and df = 35, the critical values are ±2.028.
5. Compare the test statistic with the critical values:
- If the absolute value of the calculated test statistic (|-1.8| = 1.8) is greater than the critical values (2.028), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the absolute value of the calculated test statistic (1.8) is less than the critical values (2.028). Therefore, we fail to reject the null hypothesis.
6. Conclusion:
Based on the test results, there is not enough evidence to reject the null hypothesis. Thus, we do not have sufficient evidence to conclude that the average content per bottle is significantly different from 50cl at the given significance level of 0.05.
To test the hypothesis that the average content per bottle is 50cl, we can use a one-sample t-test.
Here are the steps to perform the hypothesis test:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H₀): The average content per bottle is 50cl (μ = 50).
- Alternative hypothesis (H₁): The average content per bottle is not 50cl (μ ≠ 50).
Step 2: Determine the significance level (α):
- The significance level is not given in the question. We will assume a significance level of 0.05 (5%).
Step 3: Calculate the test statistic:
- The test statistic for a one-sample t-test formula is given by t = (x̄ - μ) / (s / √n), where:
- x̄ is the sample mean (48.5cl)
- μ is the population mean (50cl)
- s is the sample standard deviation, which is estimated from the sample: s = σ / √n, where σ is the population standard deviation (5cl)
- n is the sample size (36 bottles)
Plugging in the values, we get t = (48.5 - 50) / (5 / √36)
Step 4: Determine the critical region:
- Since the alternative hypothesis is two-sided (μ ≠ 50), we will perform a two-tailed test.
- Look up the critical t-value(s) from the t-distribution table based on the desired significance level and degrees of freedom (n - 1).
- In this case, with 36 - 1 = 35 degrees of freedom and a significance level of 0.05, the critical t-values are approximately ±2.03.
Step 5: Calculate the p-value:
- The p-value is the probability of observing a test statistic as extreme as the one calculated (t-value) under the null hypothesis.
- To calculate the p-value, we will use the t-distribution and degrees of freedom (n - 1) to find the probability of obtaining a more extreme t-value.
- Using the t-distribution table or a statistical software, we can find the p-value associated with the calculated t-value.
Step 6: Decide whether to reject or fail to reject the null hypothesis:
- Compare the p-value to the significance level.
- If p-value < α (significance level), we reject the null hypothesis.
- If p-value ≥ α (significance level), we fail to reject the null hypothesis.
Step 7: State the conclusion:
- Based on the p-value comparison, state the conclusion in the context of the problem.
Note: Since the degrees of freedom are not provided in the question, we assume a sample size of 36, which results in 35 degrees of freedom.