Question 1186469: The average score of all fourth grader in a certain certain parish on the grade 4 for numeracy is 75 with a standard deviation of 8.1 a random sample of students 100 students in one school was taken taken the mean score of 100 students is 71 does this indicate that the student of the school is significantly less skilled in the numeracy ability use a 5% level significant
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! To determine if the students in the school are significantly less skilled in numeracy ability, we can perform a hypothesis test. Here's how:
**1. State the Hypotheses**
* **Null Hypothesis (H₀):** The mean score of students in the school is equal to or greater than the parish average (μ ≥ 75).
* **Alternative Hypothesis (H₁):** The mean score of students in the school is less than the parish average (μ < 75). This is what we want to prove.
**2. Determine the Level of Significance**
* α = 0.05 (given)
**3. Calculate the Test Statistic**
We will use a one-sample z-test because we know the population standard deviation. The formula for the z-score is:
z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
z = (71 - 75) / (8.1 / sqrt(100))
z = -4.94
**4. Determine the Critical Value**
Since this is a one-tailed test (we are testing if the mean is *less* than 75), we look for the critical z-value that corresponds to α = 0.05 in the left tail of the standard normal distribution. This value is approximately -1.645.
**5. Make a Decision**
* **Compare the test statistic to the critical value:** Our calculated z-score (-4.94) is less than the critical value (-1.645).
* **Decision:** Because the test statistic falls in the rejection region (it's more extreme than the critical value), we reject the null hypothesis.
**6. Conclusion**
There is sufficient evidence at the 5% level of significance to conclude that the students in the school are significantly less skilled in numeracy ability compared to the parish average.
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