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STatiana [176]
3 years ago
12

A bank compares two proposals to increase the amount that its credit card customers charge on their cards. (The bank earns a per

centage of the amount charged, paid by the stores that accept the card). Proposal A offers to eliminate the annual fee for customers who charge $2,400 or more during the year. Proposal B offers a small percent of the total amount charged as a cash rebate at the end of the year. The bank offers each proposal to an SRS of 150 of its existing credit card customers. At the end of the year, the total amount charged by each customer is recorded. Here are the summary statistics:
Group n x s
A 150 $1,987 $392
B 150 $2,056 $413
A) Do the data show a significant difference between the mean amounts charged by customers offered the two plans?
B) The distributions of amounts charged are skewed to the right, but outliers are prevented by the limits that the bank imposes on credit balances. Do you think that skewness threatens the validity of the test? Explain.
Mathematics
1 answer:
gregori [183]3 years ago
7 0

Answer:

a) There is no  significant difference between the mean amounts charged

b) The skewness threatens the validity of the test

Step-by-step explanation:

<em>Given data:</em>

Group         n          x              s

 A              150    $1,987     $392

 B              150    $2,056    $413

n1 = n2 = 150

x1 = 1987

x2 = 2056

s1 = 392

s2 = 413

∝ = 0.05

Attached below is the detailed solution

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Answer:

Option A,

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Answered by GAUTHMATH

3 0
3 years ago
Warm-Up: Solve each equation for x.<br> 1. x + 5 = 15
eduard
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5 0
2 years ago
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Which is more 2/3 or 2/4
goblinko [34]
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6 0
3 years ago
A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe
PSYCHO15rus [73]

Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2  bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

7 0
3 years ago
rene charges $2.25 for gasoline plus $8.50 per hour for mowing lawns what is x the number of hours he has to mow lawns to earn t
Black_prince [1.1K]
2.25 + 8.50x = 50

8.50x = 50 - 2.25

8.50x = 47.75
——— ———
8.50. 8.50

x = 5.6 hours

Hope that helps :)
4 0
3 years ago
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