Question

According to American Consumer Credit Counseling statistics, credit card debt is the third largest source of household indebtedness. The following data were collected from a random sample of 12 credit cards. (See the example on page 359 of your textbook for a similar problem.) 7.444.17 5093.77586 Credit Card Balance 11.497.00 sample mean 1.931.00 rampledov 6,326.00 1,805.00 10,807.00 9.328.00 13,256.00 6.11.00 6,434.00 2,423,00 1,474.00 17,178.00 3.131125257 Quartion 2A Can you conclude the mean credit card balance is less than $2840 at e=0.05? Step 1: H: 2 2 ,840.00 N.: 1 2 ,840.00 Step 2: stat = 0 . 3.131 Step 3: Fill in row 62 if the hypothesis is one tailed. Reject H, it tcrit 1 1.796 Fill in row 66 if the hypothesis is two tailed. The smaller number must be typed first. I Reject H, if tatat Step 4: = -1.796 Step 5: Fail to Reject H, Step 6: | TIPSET T 3640 UŨ Step 7: Fill in row 76 if the Z table is appropriate P-value Fill in row 80 if the table is appropriate. The lower boundary must go T P -value Quartion 2B Construct a 98% confidence interval for the population mean credit card balance. Lower Endpoint _Upper Endpoint Question 2CW hat would the P-value be if this were a two tail hypothesis test? (Mathematically manipulate the P-value from part A to get your answer) Fill in row 32 if the table is appropriate P-value = Fill in row 96 if the table is appropriate. The lower boundary must go on the left and the upper boundary on the right. P-value

Homework problem help. Step by Step

Answer 1

Sample mean using excel function AVERAGE(), x̅ = 7444.166667

Sample standard deviation using excel function STDEV.S, s = 5093.77586

Sample size, n = 12

Q2A. 1)

Null and Alternative hypothesis:

Ho : µ ≥ 2840

H1 : µ < 2840

2)

df = n-1 = 11

Critical value :

Left tailed critical value, t-crit = T.INV(0.05, 11) = -1.796

3)

Reject Ho if t crit < -1.796

4)

Test statistic:

t = (x̅- µ)/(s/√n) = (7444.1667 - 2840)/(5093.7759/√12) = 3.1311

5)

Fail to reject Ho

6)

TINSETC µ < 2840

7)

p-value :

Left tailed p-value = T.DIST(3.1311, 11, 1) = 0.9952

0.50 < P-value <= 1.00

Q2B.

98% Confidence interval :

At α = 0.02 and df = n-1 = 11, two tailed critical value, t-crit = T.INV.2T(0.02, 11) = 2.718

Lower Bound = x̅ - t-crit*s/√n = 7444.1667 - 2.718 * 5093.7759/√12 = 3447.3768

Upper Bound = x̅ + t-crit*s/√n = 7444.1667 + 2.718 * 5093.7759/√12 = 11440.9565

3447.3768 < µ < 11440.9565

Q2C.

Two tailed p-value = T.DIST.2T(ABS(3.1311), 11) = 0.0096

0.005 < p-value <= 0.01