Using the data given below,perform a two-tailed t-test to compare the sample mean to .Use . 38 40 41 37 35 38 41
39 34 35 38 42 44 37

Question

Using the data given below,perform a two-tailed t-test to compare the sample mean to .Use . 38 40 41 37 35 38 41 39 34 35 38 42 44 37 A)2.160 < 2.601,Fail to reject B)2.601 > 2.160,Reject C)-2.160 > -2.601,Reject D)-2.601 < -2.160,Fail to reject

Quizplus · Accepted Answer

We need to perform a two-tailed t-test as we are not given any direction of inequality. The null hypothesis is that the sample mean is equal to 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00 and the alternative hypothesis is that the sample mean is not equal to 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00. 
We will use the formula t = (x̄ - μ) / (s / √n) where x̄ is the sample mean, μ is the population mean (in this case our hypothesized value of 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00), s is the sample standard deviation, and n is the sample size. 
Using the given data, we have x̄ = 38.14, μ = 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00, s = 3.06, and n = 14. 
Plugging these values into the formula gives us t = (38.14 - 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00) / (3.06 / √14) = 4.40 
We need to compare this t-value to the critical t-value with a significance level of α/2 = 0.025 and degrees of freedom (df) = n-1 = 13. Using a t-distribution table or calculator, we find the critical t-value to be 2.160. 
Since our calculated t-value of 4.40 is greater than the critical t-value of 2.160, we reject the null hypothesis and conclude that the sample mean is significantly different from 11eaab01_abc3_9590_acdb_9fd0fedd3941_TB4035_00. The correct choice is therefore B: 2.601 > 2.160, Reject 11eaab01_abc3_bca3_acdb_63ab04ea01c0_TB4035_11.

Using the Data Given Below,perform a Two-Tailed T-Test to Compare