Question 981696: A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms of their empathy for others. On a questionnaire designed to measure empathy, the mean score for the general population is 52.42. Random sampling of 51 scores obtained from individuals who were bullied yielded a mean of 99.58 and a standard deviation of 5.01. (Use these numbers only for this question!)
What is the z-value or t-value you obtained (your test statistic)? (numeric value only, including negative sign if required)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Given Info:
mu = 52.42
xbar = 99.58
s = 5.01
n = 51
n = 51 is the sample size. Because n > 30, we can use a z-score instead of a t-score. The sample size is large enough even though we don't know sigma. Also, the sample size is large enough that the central limit theorem says the distribution of sample means (xbar distribution) is approximately normal.
Z test statistic
z = (xbar - mu)/(s/sqrt(n))
z = (99.58 - 52.42)/(5.01/sqrt(51))
z = 67.2235059261639
z = 67.22
Rounded to 2 decimal places, the z-value is approximately z = 67.22
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