Question

The Nellie Mae organization found that in a random sample of undergraduate students taken in 2004, the average credit card balance was $2169. Take the sample size to be 1074, which corresponds to the 76 % of the sample of 1413 students who hold a credit card. Suppose (for now) the sample standard deviation of these credit card balances is $ 1000 a. Conduct a significance test of whether or not the sample data provide strong evidence (at the = 0.05 level) that the population mean credit card balance exceeds $ 1999. Report all aspects of the test, including a check of the technical conditions. for the lower limit. If Round the test statistic to two decimal places. Use the table to find a range for the p-value. If there is no lower limit on the p-value, enter there is no upper limit on the p-value, enter 1 for the upper limit. The technical conditions are satisfied 1999 The null hypothesis is MOUS 1999 1999 The alternative hypothesis is H Test statistiet -15.5

Answer 1

(a)

Here we have I =2169, s =1000, n=1074 Hypotheses are: H:u= 1999 H:\> 1999 Level of significance: a =0.05 Test is one tailed (right tailed). Degree of freedom: df =n-1=1073 Test Statistics: I-u t=~ = (2169-1999)/(1000/sqrt(1074)) s/n = 5.57 Critical value: Rejection Region: 1.646 If |t| >1.646, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.1,1073) =TDIST(5.57,1073,1)

From table,

The p-value is

0 < p-value < 0.0005

Since p-value is less than level of significance so we reject the null hypothesis.

(b)

Here we have i =2169, s =2000, n=1074 Hypotheses are: H :u= 1999 Hu> 1999 Level of significance: a =0.05 Test is one tailed (right tailed). Degree of freedom: df =n-1=1073 Test Statistics: T-u ta = (2169-1999)/(2000/sqrt(1074)) = 2.79 Critical value: Rejection Region: 1.646 If It| >1.646, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0.003 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.1,1073) ETDIST(2.79,1073,1)

From table,

The p-value is

0.001 < p-value < 0.005

Since p-value is less than level of significance so we reject the null hypothesis.

(c)

The second scenario produces greater p-value. This makes sense because a larger standard deviation will result in a smaller test statistics ad therefore a larger p-value.