# Conclusion do not reject null hypothesis

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ConclusionDo Not Reject Null Hypothesis"=IF(G17<B9,"Reject Null Hypothesis","DConclusion & Assumption:The conclusion is that since the p-value is > than the level of significance so we do no reject the null hyp
o Not Reject Null Hypothesis")"Do Not Reject Null Hypothesis")"T.2T(-B11,B5-1))"Do Not Reject Null Hypothesis")"pothesis, The school can reach the conclusion that the applicants average work experience is < 3 years
s.
Credit Card ChargesOne Sample Test for the MeanSample Size42Sample Mean1376.54Sample Standard Deviation183.89Hypothesized value1350Level of significance0.05t-statistic0.935335571"=(B6-B8)/(B7/SQRT(B5))One tailed t-value1.682878002"=T.INV(1-B9,B5-1)"Two tailed t-value (+ and -)2.01954097"=T.INV.2T(B9,B5-1)"Hypothesis Statement is: H0:μ = 1350H1: μ > 1350Upper tailed test
Hypothesis test resultsLower one-tailed testCritical t-value-1.68287800213271p-value0.822453989317345ConclusionDo Not Reject Null HypothesisUpper one-tailed testCritical t-value1.68287800213271p-value0.177546010682655ConclusionDo Not Reject Null HypothesisTwo-tailed testCritical t-value (+ and -)2.01954097044138p-value0.35509202136531ConclusionDo Not Reject Null HypothesisConclusion is:The conclusion is that since the p-value is > than the level of
"=-B12""=T.DIST(\$B\$11,\$B\$5-1,TRUE)""=IF(G7<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")""=B12""=1-T.DIST(\$B\$11,\$B\$5-1,TRUE)""=IF(G12<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")""=B13""=T.DIST.2T(B11,B5-1)""=IF(G17<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")"f significance so we do no reject the null hypothesis, so the data does not provide statistical evid
dence that the average monthly charges have increased.
Advertising Strategya.Hypothesis Statement is:H0 : μ < \$70.00H1: b.X_bar\$ 75.86 S.D.\$ 50.90 Mean\$ 70.00 n300t-stat1.99407028140543Critical Value1.64996576742639Decision/ConclusionReject Null Hypothesisc.X_bar\$ 68.53 S.D.\$ 45.29 Mean\$ 70.00 n700t-stat-0.858744629557285Critical Value1.64703646375354Decision/ConclusionDo Not Reject Null Hypothesisμ >\$70.00
18-25 group"=(C7-C10)/(C8/SQRT(C11))""=T.INV(0.95,C11-1)"the t-stat is > the critical value. Because of this we reject the null hypothesis. 35+ group"=(C16-C19)/(C17/SQRT(C20))""=T.INV(0.95,C20-1)"the t-stat is < the critical value. Because of this we do not reject the null hypothesis.