the process that produces sonora bars a type of candy is intended to pduce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Test to see wheather the candy bars are smaller than they are supposed to be alpha is .05 what isthe p value?
To test whether the candy bars are smaller than they are supposed to be, we can conduct a hypothesis test.
Step 1: State the hypotheses:
The null hypothesis (H0): The mean weight of the candy bars is equal to 56 gm.
The alternative hypothesis (Ha): The mean weight of the candy bars is smaller than 56 gm.
Step 2: Set the significance level (alpha):
The significance level (alpha) is given as 0.05.
Step 3: Compute the test statistic:
The test statistic for this problem is the z-score. We can calculate it using the formula:
z = (sample mean - population mean) / (population standard deviation / √sample size)
Given:
Sample mean (x̄) = 55.82 gm
Population mean (μ0) = 56 gm
Population standard deviation (σ) = 0.77 gm
Sample size (n) = 49
z = (55.82 - 56) / (0.77 / √49)
z = (55.82 - 56) / (0.77 / 7)
z ≈ -0.18
Step 4: Determine the p-value:
To find the p-value, we look up the z-score in the standard normal distribution table or use statistical software. In this case, we have a negative z-score, so we need to find the area to the left of -0.18.
Using a standard normal distribution table or statistical software, we find that the area to the left of -0.18 is approximately 0.4286.
Step 5: Compare the p-value to the significance level:
The p-value (0.4286) is greater than the significance level (0.05). Therefore, we fail to reject the null hypothesis.
Step 6: Make a conclusion:
Based on the test, there is insufficient evidence to conclude that the candy bars are smaller than they are supposed to be.
To test whether the candy bars are smaller than they are supposed to be, we can conduct a hypothesis test. Let's assume the null hypothesis (H₀) as "the candy bars are the intended size" and the alternative hypothesis (H₁) as "the candy bars are smaller than the intended size".
The given information is as follows:
- Mean weight of Sonora bars (μ) = 56 gm
- Standard deviation of the process (σ) = 0.77 gm
- Sample size (n) = 49
- Sample mean (x̄) = 55.82 gm
- Significance level (α) = 0.05 (given)
1. Calculate the standard error of the sample mean:
Standard Error (SE) = σ / √n
SE = 0.77 / √49
SE = 0.77 / 7
SE ≈ 0.11 gm
2. Calculate the z-score:
z = (x̄ - μ) / SE
z = (55.82 - 56) / 0.11
z ≈ -1.82
3. Find the p-value corresponding to the z-score using a standard normal distribution table or a statistical calculator. In this case, we are looking for the left-tail area (smaller values):
p-value ≈ 0.034
4. Compare the p-value with the significance level (α):
p-value = 0.034
α = 0.05
Since the p-value (0.034) is less than the significance level (0.05), we have evidence to reject the null hypothesis.
Therefore, the p-value for this test is approximately 0.034.