Question

5. The researcher decides to test their claim and randomly samples 50 college undergraduates and measures their outstanding credit card debt. From this sample, they calculate a mean of $3114 and a standard deviation of 146. a) Using the above data, calculate the appropriate test statistic for the original claim. b) What distribution is used to find the p-value in this situation? Be as specific as possible c) Sketch the distribution you just named, and shade the area that corresponds to the p-value. d) Based on the researchers data, the appropriate p-value for this problem is 0.003126. Give the decision at a 5% alpha level. 6. Regardless of the decision you just gave, interpret the following in the context of this problem. using plain English. You shouldn't ever use the word 'hypothesis a) A type I error b) A type II error c) The p-value of 0.003126.

Answer 1

(a). We are given that . The Null hypothesis to be tested is

I= 3114, s = 146, n = 50

$H_{0}:\mu=3173$

$H_{0}:\mu<3173$

Test statistic: $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ will have a t distribution with n-1 df.

t= 3114 - 3173 146/150

$t=\frac{-59}{146/7.0711}$

$t=\frac{-59}{20.6475}$

$t=-2.8575$

(b). A t-distribution with n-1=49 degrees of freedom is used to find out the p-value.

(c).

$normaldistributiongrapher.php?mean=0&sig$

(d). Since the p-value <0.05, we reject the null hypothesis. Hence, we conclude that there is enough evidence in the researcher's claim that college undergraduates today are better at managing their finances well, and as result have less outstanding credit-card debt currently than what was reported in 2010.

6.

(a). Type I error:

Deciding that the credit card outstanding less but actually the outstanding is more.

(b). Type II error.

Concluding that there is less outstanding on credit card bills compared to 2010 and in reality the bill is more this year.

(c). Chances of observing a difference by chance alone is 0.0031 .