-A committee at the College Board has been asked to study the SAT math scores fo
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Question
-A committee at the College Board has been asked to study the SAT math scores for students in Pennsylvania and Ohio. A sample of 45 students from Pennsylvania had an average score of 580, whereas a sample of 38 students had an average score of 530. The sample standard deviations for Pennsylvania and Ohio are 105 and 114 respectively. Does the study suggest that the SAT math score for students in Pennsylvania and Ohio differ?
a. The 95% confidence interval for the proportion is (0.4475, .8118). a. Based on the confidence interval given, is there significant evidence that the proportion of male goslings is not 50%? WHY?
b. If you had instead created a 99% confidence interval how would it compare to the one given here?
Explanation / Answer
H0:mu1-mu2=0 (there is no difference in SAT math score for students in Pennsylvania and Ohio)
H1:mu1-mu2=/=0 (there is difference in SAT math score for students in Pennsylvania and Ohio)
The assumptions are reasonable and conditions are met, use Students' t model to perform a two-sample t test for means [3(a)].
Test statistic: t=(x1bar-x2bar)/sqrt[s1^2/n1+s2^2/n2], where, xbar is sample mean, s is sample standard deviation, and n is sample size.
=(580530)/sqrt[105^2/45+114^2/38]
=2.06
Using technology, pvalue at t(76) is 0.042.
Decision rule:Reject null hypothesis if p value is less than alpha=0.05. Here, p value is les than 0.05, therefore, reject null hypothesis.
Conclusion: there is sufficient sample evidence to conclude that there is significant difference in SAT math score for students in Pennsylvania and Ohio.
a.
Hypotheses: H0:p=0.5 (proportion of male gosling is 50%)
H1:p=/=0.5 (proportion of male gosling is not 50%)
Assumptions: Model:random sampling; level of measurement:nominal, sampling distribution:normal. Both np=27*0.5=13.5 and n(1-p)=27*0.5=13.5 are atleast 5. Therefore, use one-sample Z test for proportions [3(a)].
Test statistic: z=(phat-p)/sqrt[p(1-p)/n], where, phat is sample proportion, p is population proportion, n is sample size.
=(17/27-0.5)/sqrt[0.5(1-0.5)/27]
=1.35
pvalue is 0.178.
Decsiion rule: reject null hypothesi if p value is less than alpha=0.05. Here, p value is not less than 0.05, therefore, fail to reject null hypothesis.
Conclusion: There is insufficient sample evidence to conclude that proportion of male gosling is not 50%.
Hope this will be helpful to you. Thanks and god bless you :-)
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