SOLUTION: the student has a mean of 65 and a standard deviation of 5 on a valid test and the possible score is 0 of 100, with the (alpha) of p=.01 two-tailed.

Question 1177559: the student has a mean of 65 and a standard deviation of 5 on a valid test and the possible score is 0 of 100, with the (alpha) of p=.01 two-tailed.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
It seems like you're describing a scenario where you have test scores with a known mean and standard deviation, and you want to perform a hypothesis test. Here's how we can break it down:
**Understanding the Information**
* **Mean (μ):** The average test score is 65.
* **Standard Deviation (σ):** The spread of the scores is 5. This tells us how much the scores typically deviate from the mean.
* **Possible Score Range:** Students can score between 0 and 100 on the test.
* **Alpha (α):** This is the significance level, set at 0.01. It represents the probability of rejecting the null hypothesis when it is actually true (Type I error).
* **Two-tailed:** This means we're interested in deviations both above and below the mean.
**Setting Up the Hypothesis Test**
Since we don't have a specific claim to test, let's assume we want to test whether the true population mean of the test scores is different from a certain value. Let's say we want to test if the population mean is different from 70.
1. **Null Hypothesis (H0):** The population mean is equal to 70 (μ = 70).
2. **Alternative Hypothesis (H1):** The population mean is not equal to 70 (μ ≠ 70).
**Steps to Perform the Test**
Since we know the population standard deviation, we can use a z-test.
1. **Calculate the z-score:** z = (x̄ - μ) / (σ / √n)
Where:
* x̄ is the sample mean (we would need a sample mean to actually perform the test)
* μ is the hypothesized population mean (70 in this case)
* σ is the population standard deviation (5)
* n is the sample size (we would need a sample size to perform the test)
2. **Find the critical z-values:** Since it's a two-tailed test with α = 0.01, we need to find the z-values that cut off 0.005 (0.01/2) in each tail of the standard normal distribution. Using a z-table or calculator, the critical z-values are approximately ±2.576.
3. **Compare the calculated z-score to the critical z-values:**
* If the calculated z-score falls outside the range of -2.576 to 2.576, we reject the null hypothesis.
* If the calculated z-score falls within the range of -2.576 to 2.576, we fail to reject the null hypothesis.
**Additional Considerations**
* **Sample Data:** To actually perform the hypothesis test, you would need a sample of test scores.
* **Effect Size:** You could also calculate Cohen's d to measure the effect size, which indicates the standardized difference between the sample mean and the hypothesized population mean.
**Let me know if you have a specific sample of test scores or a different value for the hypothesized population mean, and I can help you perform the complete hypothesis test.**
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