Question

The Bank of Delmarva has branches in Dover and Baltimore. The bank uses a standard credit-rating system.

Dover: n = 40, sample mean = 43.7, and sample standard deviation = 16.2.

Baltimore: n = 60, sample mean = 48.2, and sample standard deviation = 16.5.

(a) At the 0.05 significance level, test the claim that both populations have the same mean.

Original claim is made by the Answer hypothesis.
Original claim is: Mean_Dover-Mean_BaltimoreAnswer0.
Test statistic: Answer
Critical value:±Answer
P-value: Answer
Initial conclusion: We Answer the null hypothesis.
Final conclusion: There Answer enough evidence to warrant the rejection of the claim.
(b) Construct a 95% confidence interval for the difference between the two proportions.
Answer< Mean_Dover-Mean_Baltimore <Answer

Answer 1

(a)

Original claim is: Mean_Dover-Mean_Baltimore = 0.

The  standard error (SE) of the sampling distribution.

SE = sqrt[ (s₁²/n₁) + (s₂²/n₂) ]

where s₁ is the  standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

SE = sqrt[ (16.2²/40) + (16.5²/60) ] = 3.331441

Test Statistic = (x1 - x2) / SE = (43.7 - 48.2) / 3.331441 = -1.350767

The degrees of freedom (DF) is:

DF = (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

= (16.2²/40 + 16.5²/60)² / { [ (16.2² / 40)² / (40 - 1) ] + [ (16.5² / 60)² / (60 - 1) ] }

= 85 (Rounding to nearest integer)

Critical value of t at 0.05 significance level and df = 85 is 1.99

P-value = 2 * P(t < -1.35, df = 85) = 0.1806

Since p-value is greater than the 0.05 significance level,

We fail to reject the null hypothesis.

There is not enough evidence to warrant the rejection of the claim.

(b)

Margin of error = Critical t * SE = 1.99 * 3.331441 = 6.63

95% confidence interval for the difference between the two proportions

(43.7 - 48.2) - 6.63 < Mean_Dover-Mean_Baltimore <  (43.7 - 48.2) + 6.63

-11.13 < Mean_Dover-Mean_Baltimore <  2.13