Question

Assume that a simple random sample has been selected and test the given claim. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. Listed below are brain volumes in cm cubed of unrelated subjects used in a study. Use a 0.05 significance level to test the claim that the population of brain volumes has a mean equal to 1099.5 cm cubed. 962 1027 1273 1080 1071 1172 1067 1347 1100 1204

Answer 1

solt Given Sample : 962, 1027,1973,1080, 1071, 1177, 106.7, 1347, 1100, 1204 Mean at = {(x) 962 +1027 +12 73+losot 1071+11727 1067+1347 +100 +1204 11303 lo e 11303 - Sample variance = s² = {(x-x)² N- 1 (962-1130.3 )# (1027-1130.337 _- +(1804-1130.3) 10-1 124460.) 13828.9 Standard deviation =S=152 -13828.9 =117.59634348 117.5963

Hypothesis testo Null hypothesis to : H = 1099.5 Alternative hypothesis Ha :M #1099.5 Hence, this is two tailed test. = 1130.3 Ś = 117.5963 n = 10 Ceanificance level, a = 0.05 AT-M Test Statistis, tobserved = 1 1130.3 - 1099.5 117.5963/170 30.8 117.5963/3.1622 = 30.8 37.1861 0.8282 0.83 =

3 P-value at 0.05 + Statistic =0.83 level of significance, df = 9 is 0.428005 I P-value - 0.4981) Coitical value, tcritical = tayo,df ato.025 idf = 9 = 2.262 Cifrom t-table, two-tails, df-9) Rejection coiteria ; Rejection region is in direction of alternative hypothesis, Since. Ha:M #1099.5, dejection rule is : Reject Hy if Ito />ť > Since p-value >« or alternatlly 10.8315 2.262, we fail to reject the null hypothesis There is insufficient evidence to conclude that u is different from 1099.5

Conclusion: We have not enough evidence to Claim that the population brain volumes has a mean equal to 1099.5 cm3