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A psychiatrist hires you to study whether her clients self-disclosed more while sitting in an easy chair or lying down on a couch. All clients had previously agreed to allow the sessions to be videotaped for research purposes. You randomly assigned 10 clients to each condition. The third session for each client was videotaped and an independent observer counted the clients' disclosures. You reported that "clients made more disclosures when sitting in easy chairs (M = 18.20) than when lying down on a couch (M = 14.31), t(18) = 2.84, p < .05, two-tailed." Explain these results to a person who understands the t test for a single sample but knows nothing about the t test for independent means.
At the 10% level of significance, is there sufficient evidence to indicate that more books are being loaned out per day, on the average?
Find the margin of error for the 95% confidence interval used to estimate the population proportion.
Here are fifty observations on the weight of tea bags. Calculate the Standard Deviation and Variance.
Test for single Proportion - Are the data consistent with the model? Answer yes or no, and explain briefly.
Create a 99 percent confidence interval for difference in means between Line B and Line C.
Compute the variance and standard deviation of the number that will arrive within two days.
Find the probability of getting a head and rolling a six
The null hypothesis is that the distribution of mother's educational level is the same for both treatment groups. The value of the chi-square statistic is 7.422. Evaluate the P-value
Develop a 99% confidence interval for the population proportion.
a) Do a test to see if workers attitudes on this matter have changed using a 90% alpha. b) Explain to management what your findings are and what this suggests about future decisions vs work options.
A biologist has found the average weight of 12 randomly selected mud turtles to be 8.7 lb with standard deviation 3.6 lb. find a 90% confidence interval for the population mean weight of all such turtles.
The diameters of oranges in a certain orchard are normally distributed with a mean of 5.26 inches and a standard deviation of 0.500 inches.
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