Question

Next term, there are two sections of STAT 260 - Research Methods being offered. One is taught by Professor Smith and the other by Professor Jones. Last term, the class average from Professor Smith's section was higher. You want to test whether or not this difference is significant. A significant difference is one that is not likely to be a result of random variation. The scores from last year's classes are given in the table below. Test the claim that the average from Professor Smith's section was significantly different from Professor Jones' section. Use a 0.01 significance level.

(a) The claim states there is a significant difference in means (μ1 − μ2 ≠ 0). What type of test is this? This is a right-tailed test.This is a two-tailed test.     This is a left-tailed test. (b) Use software to calculate the test statistic. Do not 'pool' the variance. This means you do not assume equal variances. Round your answer to 2 decimal places. t = (c) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value =   (d) What is the conclusion regarding the null hypothesis? reject H0fail to reject H0      (e) Choose the appropriate concluding statement. The data supports the claim that the average from Professor Smith's section was significantly different from Professor Jones' section.While the average from Professor Smith's section was higher than Professor Jones', the difference was not great enough to be considered significant.      We have proven that Professor Smith is the easier professor. We have proven there was no difference between the averages from the two different professors.

Prof Smith (x1) Prof Jones (x2)    48 43 69 63 55 48 86 80 89 83 87 82 96 90 73 66 76 69 72 66 83 78 77 70 88 82 89 84 72 65 79 72 81 74 75 70 98 91 88 81 68 62 63 56 80 79 100 86 93 71

Answer 1

The step-by-step instructions to obtain Two-sample t-test using Excel 2010 function: → Type the data into the Excel2010 spread sheet. → Put the variable 1 as Sample1 and the variable 2 as Sample 2. → Click Data → Data Analysis → The t-test: Two-sample Assuming Equal Variances. → Select the variablel range. → Select the variable 2 range. → Tick Labels → The confidence level is a = 0.01 → Select the ouput range to get the required output in the spread sheet. → Click ok. The t-test: Two-sample Assuming Equal Variances :

It-Test: Two-Sample Assuming Equal Variances Mean Variance Observations Pooled Variance Hypothesized Mean Difference Sample1 Sample2 79.32143 71.59091 150.7447 158.2532 28 154.0297 48 t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail It Critical two-tail 2.18631 0.01685 2.406581 0.033701 2.682204

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(a) Let 4 and 4 be the population means and let X, and xz be the means of the respective samples. The sample statistics Sample1: = 28 x1 = 79.3 s = 12.3 Sample2:n, = 22 x2 = 71.6 s = 12.6. The level of significance is a =0.01. Step1: State the null and alternative hypotheses. The null and alternative hypotheses are, respectively. We need to use two-tailed test. H.: 44 = 4, (The two population means are equal) versus H:47H (The population mean 4 is not equal to 43) ----- ------ -- ------ ------------ ------ ----- -- Step 2: Select the distribution to use. Here, the two samples are independent, g, and , are unknown but equal, and the sample sizes are small but both populations are normally distributed Hence, all conditions are satisfied. Consequently, we will use the t distribution Step 3: Determine the rejection and non-rejection regions. The + sign in the alternative hypothesis indicates that the test is two-tailed. The significance level is 0.01. Therefore, the degrees of freedom is =+n, -2 = 28+22-2 => 50-2 =48 df = 48 --------- The critical value of t for df = 48 and 0.05 area in right-tail of the i distribution curve are 2.010634758 The critical value of t for df = 48 and 0.05 area in left-tail of the t distribution curve are -2.010634758 - - - - - - - - - - - - - - - - - - - - - - - -

(6) The standard pooled standard deviation is (n-1)=1 +(12-1) V + My - 2 (28-1)150.7447+(22-1)158.2532 28+22-2 (27)150.7447 +(21)158.2532 50-2 4070.107+3323,318 V 48 7393.425 V48 = V154.0297 = 12.41087 s, = 12.41087 Therefore, the pooled standard deviation is sy = 12.41087 - - - - - - - - - - - - - - - - - - Therefore, the standard error of the difference between the population means is --5. 171 1007 1.1 = 12.41087/28+za = 12.41087/0.035714286+0.045454545 = 12.41087J0.081169 = 12.41087(0.284901) = 3.535875 Sfi-13 = 13.535875 The test statistic is of the technology output of the part (a) is The formula for calculating the t-test statistic is (1-x) -(4-1) (79.32143-71.59091)-(0) 3.535875 7.730519 3.535875 = 2.18631 t = 2.18631 The p-value is of the technology output of the part(a) is P-value = 2P(t>to) = 2(1 - P(1 <2.18631)) = 2(1-0.98315) = 2(0.01685) = 0.033701 P-value = 0.033701 > a = 0.01 We do not reject the null hypothesis H. The critical value is t0.0148 = 2.682204 -------------------------------------------

Since, the probability value is P-value = 0.033701 > a = 0.01 P-value = 0.033701 > a = 0.01). We do not reject the null hypothesis Ho. --------------------------------------- ---- Conclusion : Since, the test statistic is kl= 2.18631 <0.0148 = 2.682204 We do not reject the null hypothesis H. Hence, we conclude that there is not significant difference between the two population means at the 1% level of significance.