) You have data on standardized economics final exams for ECON 1402 (standardize
ID: 1184853 • Letter: #
Question
) You have data on standardized economics final exams for ECON 1402 (standardized here means it is expressed as percent). You are interested in explaining how attendance and student ability affects their performance in the class. You estimate the following for 54 students: StandardFinali = 2.05 + 6.702 ln(attendance_rate)i-1.63 priorGPAi-0.128 ACTi+ui (1.36) (3.01) (0.48) (0.098) R2= 0.359 Attendance_rate is the percentage of classes attended priorGPA is the prior college grade point average ACT is the score on the ACT (a college admissions exam a la the SAT) Mean Final Exam score = 50 Mean attendance rate = 70 Mean ACT = 18 Mean GPA = 2.0 a) Interpret the coefficient on ln(attendance rate). b) Test whether the coefficient on ln(attendance rate) is different from zero against the alternative that it is positive. Please indicate your null hypothesis, alternative hypothesis, test statistic, degrees of freedom as appropriate, and result. Use a 5% significance level (the t-table is at the end of the book or else you can easily find it on-line.) c) Test whether the coefficient on priorGPA is different from zero as opposed to being equal to zero. Please indicate your null hypothesis, alternative hypothesis, test statistic, degrees of freedom as appropriate, and result. Use a 5% significance level (the t-table is at the end of the book or else you can easily find it on-line.) d) Construct the 5% confidence interval for ln(attendance rate). e) Is the coefficient on ln(attendance rate) significantly different from 8 at the 5% significance level?Explanation / Answer
what is the observed value?
ANSWER: Sample Mean = 6754
What is the test statisitc? i get this as the h0
ANSWER: t = (6754 - 7725) / 1142 = -0.85
what is the null distribution of the test statistic?
ANSWER: Null hypothesis Ho: ? = 7725
Why???
SINGLE SAMPLE TEST, "two-tailed test"
7 - Step Procedure for t Distributions
1. Parameter of interest: "?" = population mean
2. Null hypothesis Ho: ? = 7725
3. Alternative hypothesis Ha: ? ? 7725
4. Test statistic formula: t = (x-bar - ?) / s
x-bar = estimate of the Population Mean (statistical mean of the sample) [6754]
n=number of individuals in the sample [11]
s=sample standard deviation [1142]
?=Population Mean [7725]
5. Computation of Test statistic formula t = (x-bar - ?)/(s/SQRT(n))
t = (6754 - 7725) / 1142 = -0.85
6. Determination of the P-value: test based on n -1 = 10 df (degrees of freedom). Table "look-up" value shows area under 10 df curve to the left of P-value = 0.21
7. Conclusion: For 95% confidence interval (significance value ? = 0.05/2 "two-tailed"), above shows P-value > ?, [.0.21 > 0.05/2]. Do not reject Null hypothesis Ho: ? = 7725. Sample statistic does not show proof of different populatrion mean.
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