Question 1 Sample size = 200 Number of students received grades more than ‘B’= 112 Significance level = 10% Claim: More than 50% students received grades better than “B” A)Null and alternative hypothesis H0:π=0.50 Ha:π>0.50 B)Z critical value Alpha (Significance level) = 0.10 It is a right tailed hypothesis test based on the alternative hypothesis. 1
Decision rule: Null hypothesis will only be rejected when the computed z statistics is more than z critical. C)The sample proportion and z statistics p=112 200=0.56 zstatistics=p−π √(π1−π n)=0.56−0.5 √(0.5∗1−0.5 200)=1.6971 D)The p value E)Conclusion P value does not come out to be higher than the provided significance level and hence, null hypothesis would be rejected. Also, the z statistics is more than the z critical which is evidence of rejection of null hypothesis. 2
The final conclusion can be drawn that “More than 50% students received grades better than “B.” Question 2 Sample size = 25 Mean number of cups used for coffee drinking purpose by student = 4.4 Standard deviation = 0.5 Significance level = 1% Claim: On an average a student drinks 4 cups of coffee per day. A)Null and alternative hypothesis H0:μ=4 Ha:μ≠4 B)Z critical value Alpha (Significance level) = 0.01 It is a two tailed hypothesis test based on the alternative hypothesis. 3
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Decision rule: Null hypothesis will only be rejected when the computed z statistics is higher than the upper critical value or lower than the lower critical value. C)The sample proportion and z statistics zstatistics=x−μ σ √(n) =4.4−4 0.5 √(25) =4 D)The p value E)Conclusion P value is lesser than significance level and thus, null hypothesis would be rejected. Also, the z statistics is more than the upper z critical value which is evidence of rejection of null hypothesis. It can be said that on an average a student drinks 4 cups of coffee per day. Question 3 Sample 1 Who Awake 4
Sample size = 14 Mean grade = 68% Standard deviation = 8% Sample 2 Who Slept Sample size = 12 Mean grade = 75% Standard deviation = 7% Significance level = 5% Claim: Students who are stayed awake all night perform worse on an average. A)Null and alternative hypothesis H0:μA=μS Ha:μA<μS B)Degree of freedom Degree of freedom = n1+n2-2 = 14 + 12 – 2 = 24 C)t critical value Alpha (Significance level) = 0.05 Degree of freedom = 24 Critical value of t = T.INV (0.05,24) = -1.711 Decision rule: Null hypothesis will only be rejected when the computed t statistics is lower than the critical value of t in case of lower tailed hypothesis test. 5
D)The t statistics Pooled variance s2p=(n−1)s12+(n2−1)s2 n1+n2−2=(14−1)82+(12−1)72 24=57.1250 tstatistics=x1−x2 s2p(1 n1+1 n2)=68−75 57.1250(1 14+1 12)=−2.3543 E)Conclusion The calculated t statistics is lower than the t critical value which means null hypothesis will be rejected. It can be said that on an average student who stay awake all night perform worse. Question 4 Sample size = 10 Mean improvement = 0.57 Standard deviation of improvement =1.048 Significance level = 5% Claim: A)Null and alternative hypothesis H0:μD=0 Ha:μD>0 B)Degree of freedom Degree of freedom = n-1 = 10-1 = 9 6
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C)t critical value Alpha (Significance level) = 0.05 (Assuming) Degree of freedom = 9 Critical value of t = T.INV (0.05, 9) = 1.833 Decision rule: Null hypothesis will only be rejected when the computed t statistics is more than the critical value of t in case of upper tailed hypothesis test. D)The t statistics tstatistics=d√n sd=0.57 1.048∗√10=1.720 E)Conclusion The calculated t statistics (1.720) is lower than the t critical value (1.833) which means null hypothesis will not be rejected. Hence, it can be said that drug does not change the performance. 7