C Chegg Stucty | Guided Sc x \\ C i Only Need The Ones TxM WileyPLUS e - o x + e
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C Chegg Stucty | Guided Sc x C i Only Need The Ones TxM WileyPLUS e - o x + e Secure | https://edugen.wileyplus.com/edugen/student/mainfr.uni Q : # Apps D TWU Home - Texas V. Ba My Blackboard - Blac & Home Q Your Sets | Quizlet - Texas Woman's Unive ili Danielle Hardesty | Te C Chegg Study | Guide D WileyPLUS - Home / WileyPLUS: MileyPLUIS Help! I Contact Us I Log out WileyPLUS Rossman, Workshop Statistics: Discovery with Data, 4e ELEMENTARY STATISTICS I (MATH 1703/1713) Home Read, Study Practice Assignment Gradebook Assignment > Open Assignment FULL SCREEN PAINTER VERSION 1 BACK INDET ASSIGNMENT RESOURCES Unit 4, Topic 20 Unit 4, Topic 20, Exercise 20-10 a 4. Tonic 20. Credit Card Usage Exircle 20- Uit 4, Topic 20 Exercise 20-10 The Nellie Mae organization found that ina random sample of undergraduate students taken in 2004, the average credit card balance was $2169 , Take the sample size to be 1074, unch corresponds to the 76% of the sample of 1413 students Escrise 20-12 who hold credit card. Suppose (for now) the sample standard deviation of these credit card belances is $ 1000, Iait 4, Topik 20. Exercise 20-15 kai. Tokic 20. a. Conduct a significance test of whether or not the sample data provide strong evidence (as the Q = 0.05 level) that the population mean credit card balance exceeds $ 1993. Raport all aspects of the test, including a chede of the technical conditions. E 4 fue 4Tops 22 Round the test statistics to two decimal places. Use the table to find a range for the Prvalue. If there is no lower limit on the Prvalue, enter for the lower limit. If there is no upper limit on the P-value, enter for the upper limit. se 22 The technical conditions are Review Score Review. Results by Study Objective The null hypothesis is HoPT The alternative hypothesis is Hip Test statistic: (= L |Explanation / Answer
Null hypothesis is H0: mu = 1993
Alternate hypothesis is H1: mu > 1993
Test statistics, t = (2169 - 1993)/(1000/sqrt(1074)) = 5.7678
0 < p-value < 0.05
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