In anthropological studies, an important characteristic of fossils is cranial capacity. Frequently skulls are at least partially decomposed, so it isnecessary to use other characteristics to obtain information about cranial capacity. One measure that has been used is the length of the lambda-opisthion chord. An article reported the accompanying data for n = 7 Homo erectus fossils.78 74 78 82x (chord length in mm)y (capacity in cm³)Suppose that from previous evidence, anthropologists had believed that for each 1 mm increase in chord length, cranial capacity would be expectedto increase by 20 cm³. Do the data given here strongly contradict prior belief? Use a 0.05 level of significance.Calculate the test statistic. (Round your answer to two decimal places.)t =83850 77586750 975 915881,015 1,030Use technology or the Distribution Calculators page in SALT to find the P-value for this test. (Round your answer to four decimal places.)P-value =What can you conclude?O Fail to reject Ho. We do not have convincing evidence that the increase in cranial capacity associated with a 1 mm increase in chord length isnot 20 cm ³O Reject Ho. We have convincing evidence that the increase in cranial capacity associated with a 1 mm increase in chord length is not 20cm ³.Fail to reject Ho. We have convincing evidence that the increase in cranial capacity associated with a 1 mm increase in chord length is not 20cm 3O Reject Ho. We do not have convincing evidence that the increase in cranial capacity associated with a 1 mm increase in chord length is not 203cm.

Given information, x y 78 850 74 775 78 750 82 975 83 915 86 1015 88 1030Using the provided data, the regression equation is obtained as follows: y^=a+bx b=∑(x-x¯)(y-y¯)∑(x-x¯)2a=∑y-b∑xn Using the provided data, the below table is obtained. s.no x y (x-x¯) (y-y¯) (x-x¯)2 (y-y¯)2 (x-x¯)(y-y¯) 1 78 850 -3.2857 -51.4286 10.79582 2644.901 168.979 2 74 775 -7.2857 -126.429 53.08142 15984.19 921.1209 3 78 750 -3.2857 -151.429 10.79582 22930.62 497.549 4 82 975 0.7143 73.5714 0.510224 5412.751 52.55205 5 83 915 1.7143 13.5714 2.938824 184.1829 23.26545 6 86 1015 4.7143 113.5714 22.22462 12898.46 535.4097 7 88 1030 6.7143 128.5714 45.08182 16530.6 863.267 Total 569 6310 145.4286 76585.71 3062.143 x¯=∑xin=78+74+78+82+83+86+887=5697=81.2857 y¯=∑yin=850+775+750+975+915+1015+10307=63107=901.4286 Using the table values, the regression coefficients are calculated as follows: b=∑(x-x¯)(y-y¯)∑(x-x¯)2=3062.143145.4286=21.056 a=∑y-b∑xn=6310-(21.056)5697=6310-11980.867=-5670.867=-810.123 Using the regression coefficients, the regression equation is obtained as follows: y^=a+bxy^=-810.123+21.056x The standard error of slope formula is given as follows: sb1=y-y^2n-2∑x-x¯2 Using the regression equation, the estimated value is obtained as follows: The regression equation is y^=-810.123+21.056x s.no x y y^ (y-y^) (y-y^)2 1 78 850 832.245 17.755 315.24 2 74 775 748.021 26.979 727.8664 3 78 750 832.245 -82.245 6764.24 4 82 975 916.469 58.531 3425.878 5 83 915 937.525 -22.525 507.3756 6 86 1015 1000.693 14.307 204.6902 7 88 1030 1042.805 -12.805 163.968 Total 569 6310 12109.26 The standard error of slope is calculated as follows: sb1=y-y^2n-2∑x-x¯2=12109.267-2145.4286=49.2123112.05938=4.081 Thus, the standard error of slope is 4.081.Test statistic for provided data is calculated as follows: t=b1sb1=21.0564.081=5.1595≈5.16 Thus, the value of test statistic is 5.16. .The required p-value is obtained as follows: Degrees of freedom is required for obtaining p-value, so degrees of freedom is obtained as follows: df=n-2=7-2=5 Thus, the degrees of freedom is 5. Computation of P-value: The two tailed P-value for the t-test at 5 degrees of freedom can be obtained using the excel formula, “=T.DIST.2T(5.16,5)”. Thus, the p-value is 0.0036. Decision rule: If p-value ≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: Here, p-value (0.0036) is less than the level of significance (0.05). Thus, reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the increase in cranial capacity associated with a 1 mm increase in chord length is not 20 cm3 Thus, the correct option is, Reject H0. We have convincing evidence that the increase in cranial capacity associated with a 1 mm increase in chord length is not 20 cm3