A set of physics exam scores are normally distributed with a mean of 80 points and a standard deviation of 6
1 answer:
Answer:
The proportion of exam scores that are lower than Irina's score is 0.9938
Step-by-step explanation:
<u>Data</u>
- mean: 80 points
- standard deviation (sd): 6 points
- value of interest (x): 95 points
z-score is computed as follows:
z = (x - mean)/sd
z = (95 - 80)/6 = 2.5
We want to find:
P(z < 2.5)
From the table attached:
P(z < 2.5) = 0.9938
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Answer:
D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.
Step-by-step explanation:
We will test the hypothesis on the difference between means.
We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).
Then, the difference between means is:
If the standard deviation of the differences of the sample means of the two groups was 1.1 days, the t-statistic can be calculated as:
The critical value for a two tailed test with confidence of 95% (level of significance of 0.05) is t=z=1.96, assuming a large sample.
This is approximately 2 standards deviation (z=2).
The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.