Question

The σ population standard deviation is unknown. You collected a simple random sample of the cents portions from 100 checks and from 100 credit card charges. The cents portions of the checks have a mean of 23.8 cents and a standard deviation of 32.0 cents. The cents portions of the credit charges have a mean of 47.6 cents and a standard deviation of 33.5 cents. Use an α = 0.05 level to test the claim that the cents portions of checks have a mean less than that of credit card charges.

Using a TI 83/84 calculator, calculate my P-value with the appropriate Hypothesis Test.                                  

my P-value = _______

A. P-value is about 0.9999 or 1

B. P-value is about 0 (3.33 E-7) zero

C. P-value is about 0.238

D. P-value is about 0.476

Answer 1

. Use the TI-83 steps to find the p-value as follows: Press the STAT button and highlight Tests Press number 4:2-sample Ttest Highlight Stats and Enter the values of = 23.8, s, = 32.0,n = 100,72 = 47.6,5, = 33.5, n = 100 Selct left tailed (1 <uz) and no for pooled Highlight calculate and hit enter Output: . . TEXAS INSTRUMENTS TI-83 Plus 2-SamPTTest 1<h2 t=-5.137322183 P=3.3314541 E-7 df=197. 5859426 X1=23.8 +2=47.8 The required p-value is (3.3314541 E-7) Hence, the correct answer is option (B) B. P-value is about 0 (3.33 E-7) zero