The web based company Oh Baby! has a goal of processing 95% of its orders on the same day they are received. If 485 out of the next 500 orders are processed on the same day, would this prove that they are exceeding their goal, using a = .025?
Null hypothesis:
Ho: p = .95 -->meaning: population proportion is equal to .95
Alternative hypothesis:
Ha: p > .95 -->meaning: population proportion is greater than .95
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .97 - .95 -->test value (485/500 = .97) minus population value (.95) divided by
√[(.95)(.05)/500] -->.05 is (1 - .95) and 500 is the sample size.
Finish the calculation.
Using a z-table, find your critical or cutoff value using .025 level of significance for a one-tailed test. The test is one-tailed because the alternative hypothesis is showing a specific direction (greater than).
If the test statistic exceeds the critical or cutoff value from the table, the null is rejected in favor of the alternative hypothesis and p > .95 (there is enough evidence to prove they are exceeding their goal).
Hope this will help get you started.
Thank you for your help.
To determine whether Oh Baby! is exceeding its goal, we need to perform a hypothesis test using the given information and the significance level (α) of 0.025.
Let's set up the null (H0) and alternative (H1) hypotheses:
H0: Oh Baby! is meeting its goal of processing 95% of orders on the same day.
H1: Oh Baby! is exceeding its goal of processing 95% of orders on the same day.
Now, let's calculate the critical value for the given significance level of α = 0.025. Since the sample is large (n > 30), we can use the normal distribution.
To find the critical z-value, we can use a Z-table or a statistical calculator. A significance level of 0.025 corresponds to a two-tailed test (0.025 in each tail). Half of that probability in each tail is 0.025 / 2 = 0.0125.
Using a Z-table or a calculator, we find that the critical z-value for a significance level of 0.0125 in each tail is approximately ±1.96.
Now, let's calculate the sample proportion:
Sample Proportion (p̂) = Number of orders processed on the same day / Total number of orders = 485/500 = 0.97
To perform the hypothesis test, we need to find the test statistic (z-statistic):
z = (p̂ - p) / √((p * (1 - p)) / n)
Here,
p is the null hypothesis proportion (95% or 0.95),
p̂ is the sample proportion (0.97),
n is the sample size (500).
Let's plug the values into the formula and calculate the z-statistic:
z = (0.97 - 0.95) / √((0.95 * (1 - 0.95)) / 500)
z = 0.02 / √((0.95 * 0.05) / 500)
z ≈ 0.02 / 0.01414 ≈ 1.414
Since this is a two-tailed test, we need to compare the absolute value of the z-statistic (1.414) with the critical z-value (±1.96).
Since 1.414 < 1.96, we fail to reject the null hypothesis (H0).
Therefore, there is not enough statistical evidence to conclude that Oh Baby! is exceeding its goal of processing 95% of orders on the same day based on the given sample of 485 out of 500 orders.
Note: Statistical analysis may vary based on sample assumptions and methodology used.