demostrate steps to get to answers 202020211689.4 1848.11689.3 1876.71758.7 1877.31740.6 1890.71767.91776.81751.51742.7 1907.71708.9 1911.11721.3 1906.71698.2 1913.11891188618941749.8 1906.751757.8 1911.151748.9 1912.51738.8 1917.21738.7 1880.51730.3 1899.21713.2 1906.11721.1 1901.81741 1902.91743.8 1903.81721.7 1886.91761.1 1873.21750.3 1863.71734.4 1868.71741.717821750.31776.51774.4 1790.51790.417851772.61791.11785.11784.41792.11785.51760.51788.317711774.9730.651779.3adySheet1Scenario: You are given Gold price data from 2020 and 2021. Answer the following questions using this data.Number123QuestionsReport the mean for both years20202021Report the variance for both years20202021Question 1Accessibility: Investigate135 52,652Is there a difference in the means between 2020 and 2021?What is the null hypothesis that you need to test to answer this question?What is the test statistic you need to answer this question (what kind of statistics, not the number but the name).What is the p-value for this test?Is there a difference in the means between 2020 and 2021?Question 2Question 3DECLO5Question 4Question 5tv+AOAAnswerWW29.B?IHTMLPNGEL20

The data set is given as follows:202020211689.41848.11689.31876.71758.71877.31740.61890.71767.918911776.818861751.518941742.71907.71708.91911.11721.31906.71698.21913.11749.81906.751757.81911.151748.91912.51738.81917.21738.71880.51730.31899.21713.21906.11721.11901.817411902.91743.81903.81721.71886.91761.11873.21750.31863.71734.41868.71741.717821750.31776.51774.41790.51790.417851772.61791.11785.11784.41792.11785.51760.51788.31774.91771730.651779.3Part (1)For year 2020:The sampled number of year, n1=35.The sample average, x¯1=∑ixin=1689.4+1689.3+...+730.6535≈1716.252857For year 2021:The sample average,x¯2=1848.1+1876.7+...+1779.335=1862.011429Part (2)The sample variance for year 2020:s12=∑i(xi−x¯)2n−1=(1689.4−1716.252857)2+(1689.3−1716.252857)2+...+(730.65−1716.252857)235−1≈30110.20308.The sample variance for year 2021: s22=(1848.1−1862.011429)2+(1876.7−1862.011429)2+...+(1779.3−1862.011429)235−1≈2785.434424.Part (3)Considering the μ1 and μ2 as the true averages for the years 2020 and 2021, it is set up as follows:H0:μ1=μ2; the averages for the years 2020 and 2021 are hypothesized to be indifferent.H1:μ1≠μ2; the averages for the years 2020 and 2021 are hypothesized to be not indifferent.As the ratio of the sample variances (higher to smaller), s12s22=30110.203082785.434424≈10.81 higher than 4, using the thumb rule, the variances for two years are considered to be unequal at the respective population levels. Thus, a t-test for unequal variances is to be formulated. Test statistic:t=x¯1−x¯2s12n1+s22n2=1716.252857−1862.01142930110.2030835+2785.43442435≈−4.754The distribution degrees of freedom,df=(s12n1+s22n2)2(s12n1)2n1−1+(s22n2)2n2−1=(30110.2030835+2785.43442435)2(30110.2030835)235−1+(2785.43442435)235−1≈40The p-value 9two-tailed) for the t score of −4.754 with distribution df=40 is obtained using the excel function "T.DIST.2T(ABS(t),df) as follows:The p-value is 0.0000.The decision rule is: reject H0, "p-value≤α".For any significance level of α=0.01, 0.05, or 0.10, p-value(=0.0000)<α. Thus H0 is rejected.The averages for the years 2020 and 2021 are concluded to be not indifferent.(1) The sample averages are as follows:year 2020=1716.252857 year 2021=1862.011429(2) The sample variances are as follows:year 2020=30110.20308year 2021=2785.434424(3) The hypotheses are: H0:μ1=μ2; H1:μ1≠μ2.A t-test for unequal variances is considered.The p-value is 0.0000.Yes, a significant difference between the averages for the years 2020 and 2021 is concluded.