a professor was curious to whether the students in a very large class she was teaching who turned in their 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 score were approximately normally distributed.The mean score for the first 20 tests was 78.Did the student turning in their tests first score significantly different from the mean at the .05 level.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.
yes
To determine if the students turning in their tests first scored significantly different from the mean at the 0.05 level, we need to perform a hypothesis test.
Let's set up the null and alternative hypotheses:
- Null Hypothesis (H₀): The mean score of the students turning in their tests first is equal to the overall mean score of 75.
- Alternative Hypothesis (H₁): The mean score of the students turning in their tests first is significantly different from the overall mean score of 75.
Next, we need to calculate the test statistic. Since we know the sample mean (78), the population mean (75), and the standard deviation (10), we can use the formula for a z-test:
z = (sample mean - population mean) / (standard deviation / √sample size)
Using the given information, we can calculate the test statistic:
z = (78 - 75) / (10 / √20)
z = 3 / (10 / 4.47)
z = 1.34
Now, we need to find the critical value for a two-tailed test at a significance level of 0.05. Since the significance level is divided equally between the two tails, each tail will have an area of 0.025.
To find the critical value, we can look up the corresponding z-score from a standard normal distribution table or use a calculator. The critical value for a two-tailed test with a significance level of 0.05 is approximately ±1.96.
Since the test statistic of 1.34 falls within the range of -1.96 to +1.96, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean score of the students turning in their tests first is significantly different from the overall mean score at the 0.05 level.
In summary, based on the given information and the hypothesis test, the students turning in their tests first did not score significantly different from the overall mean at the 0.05 level.