Hello I need help with the this question. If possible figure the question in exc

Question

Hello

I need help with the this question. If possible figure the question in excel, StatCrunch or TI84.

A random sample of students at a college reported what they believed to be their heights in inches. Then the students measured each others' heights in centimeters, without shoes. The data provided are for the men, with their believed heights converted trom inches to centimeters. Assume that conditions for t-tests hold. Complete parts a and b below Click the icon to iew the data. a Find a 95% confidence interval for the mean difference as measured in centimeters. Does It capture 07 What does that show? The 95% confidence interval is 1-2107 . 0.589) (Round to three decimal places as needed) The interval does include 0, so a hypothesis that the means are equal cannot be rejected. b. Perform a ttest to test the hypothesis that the means are not the same. Use a signincance level of 0.05 Delermine the typotheses for this test Let i difference be the populaion mean difference between measured and believed height in centimeters. Choose the correct answer below. O B. Find the test slatistic for this test -1.21 (Round to two decimal places as needed.) Find the p-value for this test p-value 0.248 (Round to three decimal places as needed.) What is the concdusion for this test? OA. RejectHo The means of measured and believed heights are significantly difTerent O B. Do not reject Ho . The means of measured and believed heights are significantly different. C. Do not reject Ho. The means of measured and believed heights are not significantly different. RejectHo. The means of measured and believed heights are not significanty different

Explanation / Answer

H0: 1 - 2 = 0 i.e. (1 = 2)

H1: 1 - 2 0 i.e. (1 2)

            

Assuming population variances are equal, we would have to calculate pooled-variance t-Test

Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)

         = (15-1)*8.07^2^2+(15-1)*8.27^2/14+14

         = 911.73+957.41744/28

         = 66.7553

tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)

       =(176.53-177.292)-0/66.7554(1/15+1/15)

      =-0.2543

The P-Value is 0.400676. The result is not significant at p < .05.

Confidence interval:-

(X1-X2)+-ta/2*Sp^2(1/n1+1/n2)

-0.7586667+/-2.05*2.98

LCL=-0.7586667-6.109=-6.867

UCL=-0.7586667+6.109=5.35

(-6.867,5.35)

Centimeters (measured) Centimeters (believed) 164 165.1 170 170.18 186 185.42 167 170.18 189 193.04 174 175.26 175 177.8 189 193.04 179 175.26 177 177.8 183 180.34 174 175.26 170 172.72 168 167.64 183 180.34 176.5333333 177.292 Average 8.069932436 8.269640863 Std Dev

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