dole pineapple inc. is concerned that the 16 ounce cans are overfilled. assume the standard deviation is .03 onces. a random sample of 50 cans has a mean weight of 16.05 ounces. at the 5% level of significance can we conclude that the mean weight is > than 16 ounces. what is the p value?
Try a one-sample z-test.
Hypotheses:
Ho: µ ≤ 16 -->null hypothesis
Ha: µ > 16 -->alternate hypothesis
Data:
n = 50
sd = .03
mean = 16.05
Once you calculate z using a z-test, find the p-value by using a z-table. The p-value is the actual level of the test statistic. Compare that value to .05 to either reject the null or fail to reject the null. If you reject the null, you can conclude that the mean weight is greater than 16 ounces.
I hope this will help get you started.
To check if we can conclude that the mean weight is greater than 16 ounces, we can perform a hypothesis test. We will use a one-sample t-test in this case.
Here are the steps to perform the hypothesis test:
Step 1: Define the null hypothesis (H0) and alternative hypothesis (Ha):
- Null hypothesis (H0): The mean weight is 16 ounces.
- Alternative hypothesis (Ha): The mean weight is greater than 16 ounces.
Step 2: Determine the significance level (α):
- The significance level is given as 5%, which is equal to 0.05.
Step 3: Calculate the test statistic:
- The test statistic for a one-sample t-test can be calculated using the formula:
t = (sample mean - hypothesized mean) / (standard deviation / √sample size)
In this case, the sample mean is 16.05 ounces, the hypothesized mean is 16 ounces, the standard deviation is 0.03 ounces, and the sample size is 50. Plugging these values into the formula, we get:
t = (16.05 - 16) / (0.03 / √50)
Step 4: Determine the critical value (tcrit):
- Since the alternative hypothesis states that the mean weight is greater than 16 ounces, this is a right-tailed test.
- We will find the critical value using the t-distribution table or a statistical software.
- For a 5% level of significance and 49 degrees of freedom (sample size - 1 = 50 - 1 = 49), the critical value is approximately 1.68.
Step 5: Compare the test statistic with the critical value:
- If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, if t > tcrit (1.68), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Calculate the p-value:
- The p-value is the probability of obtaining a test statistic as extreme as the one observed in the sample, assuming the null hypothesis is true.
- To calculate the p-value, we can use the t-distribution table or a statistical software.
- In this case, since the alternative hypothesis is one-tailed (greater), we will find the area to the right of the test statistic.
- The p-value can be interpreted as the probability of observing a sample mean of 16.05 ounces or greater, assuming the population mean is 16 ounces.
Given the values from Step 3, we can calculate the test statistic (t):
t = (16.05 - 16) / (0.03 / √50)
Now we need to compare the test statistic with the critical value:
If t > tcrit (1.68), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Finally, we can calculate the p-value to determine its significance.