Question 1202041: Researchers were surprised to find that young children under stress actually report fewer symptoms of anxiety and depression than we would expect. But they also noticed that their scores on a Lie scale (a measure of the tendency to give socially desirable answers) were higher than expected. The population mean for the Lie scale on the Children’s Manifest Anxiety Scale is known to be 3.87. For a sample of 36 children under stress, the researchers found a sample mean of 4.39, with a standard deviation of 2.61. Does this group of children show an increased tendency to give socially acceptable answers? Test the hypothesis at α = 0.05. Also, calculate the 95% confidence limits and check whether the limits are consistent with your conclusion.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean 3.87
sample mean is 4.39
sample stndard deviation is 2.61
sample size is 36
since you are looking for the mean of a sample, use the standard error rather than the standard devition.
since the standard deviation is from the sample rather than the population, use the t-score rather than the z-score.
standard error = standard deviation / square root of sample size = 2.61 / sqrt(36) = .435
t-score = (x-m)/s = (4.39-3.87)/.435 = 1.195402299.
critical t-score for one tailed 95% confidence interval at 35 degrees of freedom is equal to t = 1.689572394.
since the test t-score is not greater than the critical t-score, the results are not significant.
this means there is not enough evidence to refute the claim that this group of children show an increased tendency to give socially acceptable answers.
the test alpha was equal to .1199834109.
this was greater than the critical alpha of .05, leading to the same conclusion, as it should have.
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