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Psy 315 Week 3 Agression Analysis Paper

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Question 1
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
1. Formulate the hypotheses that can be used to determine whether the mean of all account balances is significantly different from $1,150.
Solution: Null hypothesis Hₒ: µ = $ 1150 against alternative H₁: µ≠$1150

2. Compute the test statistic.

Solution: since the sample size 81>30 therefore we use Z test and the test statistics will be
Z= sample standard deviation /√ sample ¿¿ sample mean − µ ¿

Under the null hypothesis the test statistics Z= 1200 – 1150 126 / √ 81 = 7.14

3. Using the p-value …show more content…

Sample 1 size 40 Sample 2 size 45 Sample 1 mean 72 Sample 2 mean 78 Sample 1 S.D. 6 Sample 2 S.D. 8 Degree of freedom = [s1^2/n1+s2^2/n2] ^2 / [(s1^2/n1) ^2/(n1-1) +(s2^2/n2) ^2/(n2-1)]
= [36/40 + 64/45] ^2 / [(36/40) ^2 /39+ (64/45) ^2/44] Degree of freedom 80.8018 Standard error of the difference between means = sqrt (s1^2/n1+s2^2/n2) = sqrt (6^2/40+8^2/45) = 1.523884 Confidence interval: (xbar1-xbar2) - t SE and (xbar1-xbar2) + t SE t (0.05,81) = 1.99 (72-78) - (1.99) (1.523884) and (72-78) + (1.99) (1.523884) 95% confidence interval is: (-9.03, -2.97)

2. At 95% confidence, use the p-value approach and test to determine if the average yearly income of marketing managers in the East is significantly different from the West. t = [ (x bar 1 – x bar 2) - d] / SE = -6/1.523884
Test statistic t = -3.937 (df = 80), p-value < .005, reject Hₒ and conclude that there is a significant

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