Pt1420 Unit 2 Population Analysis

#14 – A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

H0: 1 = 2 H1: 1 2

n1 = 15 n2 = 17 (sample size)
1 = 350 2 = 342 (sample mean) s1 = 12 s2 = 15 (standard deviation) = 0.10 (alpha significance level)

Step 1 – State the Null and Alternative Hypothesis

H0: 1 = 2 H1: 1 2

Step 2 – Select the Significance Level

= 0.10

Step 3 = n is < than 30, so we use "T" Test Statistics

Step 4 = Find the Critical …show more content…
Otherwise do not reject H0. (formula used // =IF(1.65>2.75,"reject H0","Do not reject H0")

Based on the decision: Do not reject …show more content…
Critical Value = 1.795884 // Used table for formula t = -2.200928 // Used table for formula

Step 5 = Solve –Test Statistic greater than Critical Value then reject H0. Otherwise do not reject H0. (formula used // =IF(-2.20093>1.795884,"Reject H0","Do Not Reject H0")

Based on the decision: Do Not Reject H0.

For "P" Value = Solve – If P-value < Then reject Ho. Otherwise do not reject HO. (formula used // = =IF(0.025002469 than 30, so we use "Z" Test Statistics

Step 4 = Find the Critical Value of "Z" statistic – use the Two Sided Test).

Critical Value = 1.959961 // Formula =NORMSINV(1-0.05/2) Z= = -1161895 // Formula
=(150000-180000)/SQRT(0.05/75)+(0.05/120)

Step 5 = Solve –Absolute Value of Test Statistic greater than Critical Value then reject H0. Otherwise do not reject H0. (formula used // =IF(-2.20093>1.795884,"Reject H0","Do Not Reject H0")

Based on the decision: Do Not Reject