POW_Credit Card USe

xlsx

School

College of San Mateo *

*We aren’t endorsed by this school

Course

123

Subject

Economics

Date

May 25, 2024

Type

xlsx

Pages

3

Report

Uploaded by GeneralWillpowerAntelope75

Credit Card Use by Undergraduates. In a study entitled How Underg $3173. This figure was an all-time high and had increased 44% over t mean credit card balance for undergraduate students has continue On the basis of above information, answer the following questions: Qs. 1 State the null and alternative hypotheses. Hint: Will it be a upper one tailed, lower one tailed or a two tailed te Qs. 2 What is the p-value for a sample of 180 undergraduate studen Hint: You are given the population mean and standard deviation in th population mean)/population standard deviation/square root of the The relevant p-value calculation will depend on the alternative hypo Qs. 3 Using a .05 level of significance, what is your conclusion? Hint: Use Reject the null hypothesis if p-value< level of significance, Answers: Qs.1 Null hypothesis <= (the mean should not be more than 3173) Alternative hypothesis >3173 (if the mean is more than 3173, it'll be This is an upper one tailed test. Qs.2 P-value = 0.97928965 first calculate the z-value using the formula (sample mean - populati use =NORM.S.DIST(z-value, true) to find the probability of the lower Take 1-0.97928965 which is 0.02071035 Qs 3. The conclusion is we reject the null as the p-value is 0.02071035 and
graduate Students Use Credit Cards, Sallie Mae reported that undergr the previous five years. Assume that a current study is being conducte ed to increase compared to the April 2009 report. Based on previous s est? Your research hypothesis should be in the alternative hypothesis nts with a sample mean credit card balance of $3325? he question set up. For this part, you also are given sample mean and e sample size. othesis--upper tailed value, lower tailed p-value and two tailed p-value otherwise do not reject the null. e an alternative hypothesis which is also the esearch hypothesis) tion mean) / sample standard deviation which is 2.039294. Then r tail, which is 0.97928965 . To find the probability of upper tail, d it's smaller than the level of signinficance.
Z-value sample mean - population mean 152 sample standard deviation 74.5355992 z value 2.039294 (sample mean-population mean) / sample standard deviation P-value of lower tail 0.97928965 P-value of upper tail 0.02071035 raduate students have a mean credit card balance of ed to determine whether it can be concluded that the studies, use a population standard deviation σ=$1000 . s. d sample size. Compute the z-value= (sample mean - e should be calculated using NORM.S.DIST(z-value, TRUE)
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