Chicken Delight claims that 90 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the .10 significance level, can we conclude that less than 90 percent of the orders are delivered in less than 10 minutes?
no idea when it come to statisitics how to use the formulas correctly
any help is greatly appreciated
Let's try a binomial proportion one-sample z-test for this problem. I'll give you the setup for the calculations and let you take it from there.
Formula with your data included:
z = (.82 - .9)/√[(.9)(.1)/100]
Note: .82 is 82/100; .9 is 90% from the problem; .1 is 1-.9; 100 is the sample size.
Finish the calculation. Check a z-table for .10 level of significance to determine your critical or cutoff value to reject the null or fail to reject the null hypothesis. You can then draw your conclusions.
I hope this will help get you started.
To determine if we can conclude that less than 90 percent of the orders are delivered in less than 10 minutes, we can perform a hypothesis test.
Here's how to do it step by step:
Step 1: Formulate the hypotheses.
Null hypothesis (H0): 90% of orders are delivered within 10 minutes.
Alternative hypothesis (H1): Less than 90% of orders are delivered within 10 minutes.
Step 2: Determine the test statistic and the critical value.
Since we have a sample proportion (82 out of 100 orders), we can use the normal distribution approximation for the test statistic.
The test statistic is calculated as:
z = (p̂ - p) / √(p(1-p) / n)
Where:
p̂ is the sample proportion (82/100 = 0.82)
p is the hypothesized proportion (0.90)
n is the sample size (100)
We want to test if the sample proportion is significantly less than the hypothesized proportion, so we calculate the lower tail critical value. At a 0.10 significance level, the critical value is -1.28 (obtained from the standard normal distribution table).
Step 3: Perform the calculation.
Calculate the test statistic:
z = (0.82 - 0.90) / √(0.90 * (1-0.90) / 100)
z = -1.44
Step 4: Make the decision.
Compare the test statistic with the critical value.
Since -1.44 > -1.28, we do not reject the null hypothesis.
Step 5: Interpret the result.
Based on the calculated test statistic, we do not have sufficient evidence to conclude that less than 90% of the orders are delivered within 10 minutes, at a 0.10 significance level.
However, it is important to note that this conclusion is specific to the sample and may not necessarily represent the entire population of orders from Chicken Delight.
To test whether we can conclude that less than 90 percent of the orders are delivered in less than 10 minutes, we can set up the following hypotheses:
Null hypothesis (H0): p = 0.90
Alternate hypothesis (Ha): p < 0.90
Where p is the proportion of orders delivered within 10 minutes.
To perform the hypothesis test, we can use the z-test for proportions.
1. Calculate the test statistic (z-score):
z = (p̂ - p) / √(p * (1-p) / n)
Where p̂ is the sample proportion (82 / 100), p is the hypothesized proportion (0.90), and n is the sample size (100).
z = ((82 / 100) - 0.90) / √(0.90 * (1-0.90) / 100)
2. Find the critical z-value for a one-tailed test at a significance level of 0.10. This critical z-value represents the cutoff point beyond which we would reject the null hypothesis.
The critical z-value can be found using a z-table or a calculator and is approximately -1.28 for a one-tailed test at a significance level of 0.10.
3. Compare the test statistic (z-score) with the critical z-value:
If the test statistic is less than the critical z-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
If z < -1.28, we reject the null hypothesis.
If z >= -1.28, we fail to reject the null hypothesis.
4. Calculate the p-value associated with the test statistic:
The p-value represents the probability of observing a test statistic as extreme as the one calculated (or more extreme) assuming the null hypothesis is true.
For a one-tailed test, the p-value is the probability of observing a z-score less than or equal to the test statistic. This can be found using a z-table or a calculator.
If the p-value is less than the significance level (0.10), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Therefore, to complete the hypothesis test, calculate the test statistic (z-score), find the critical z-value, compare the test statistic with the critical value, and calculate the p-value.