A nursing professor was curious as to whether the students in a very large class she was teaching who turned in their tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 tests was 78. Did the students turning in their tests first score significantly different from the mean at the .05 level?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
0.3
To determine if the students who turned in their tests first scored significantly different from the mean at the 0.05 level, we can conduct a hypothesis test.
Here's how you can go about solving this problem step-by-step:
Step 1: Determine the null and alternative hypotheses:
- Null hypothesis (H0): The mean score of the students who turned in their tests first is equal to the overall mean score (μ = 75).
- Alternative hypothesis (Ha): The mean score of the students who turned in their tests first is significantly different from the overall mean score (μ ≠ 75).
Step 2: Set the level of significance:
The given level of significance is 0.05 or 5%.
Step 3: Calculate the test statistic:
To calculate the test statistic, we'll use the formula for a z-score, which is given by:
z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x̄ (sample mean) = 78, μ (population mean) = 75, σ (population standard deviation) = 10, and n (sample size) = 20.
Plugging in the values, we get:
z = (78 - 75) / (10 / √20)
z ≈ 1.58
Step 4: Determine the critical value(s):
Since we have a two-tailed test (μ ≠ 75), we need to find the critical value(s) for α = 0.05.
The critical value(s) can be found using a standard normal distribution table or a statistical software. For a two-tailed test at α = 0.05, the critical values are approximately ±1.96.
Step 5: Compare the test statistic with the critical value(s):
Since the test statistic (1.58) does not exceed the critical value (1.96), we fail to reject the null hypothesis.
Step 6: Draw a conclusion:
Based on the test results, we do not have sufficient evidence to conclude that the students who turned in their tests first scored significantly differently from the overall mean at the 0.05 level.
In summary, the students turning in their tests first did not score significantly different from the overall mean at the 0.05 level, based on the given data.