The average expenditure on valentine's day was expected to be$100.89 (usa today, february 13, 2006). do male and femaleconsumers differ in the amounts they spend? the average expenditurein a sample survey of 40 male consumers was $135.67, and theaverage expenditure in a sample survey of 30 female consumers was$68.64. based on past surveys, the standard deviation for maleconsumers is assumed to be $35, and the standard deviation forfemale consumers is assumed to be $20. the z value is 2.576 .

round your answers to 2 decimal places.

a. what is the point estimate of thedifference between the population mean expenditure for males andthe population mean expenditure for females?


b. at 99% confidence, what is the marginof error?


c. develop a 99% confidence interval forthe difference between the two population means.
to

(a) $62.16

(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

Confidence interval for female expenditure is ($49.18, $69.30)

Step-by-step explanation:

(a) Male expenditure

Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

Error margin = (2.708×35)/√40 = 94.78/6.32 = $15.00

Female

Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

-14 is the answer

hope this : ))

step-by-step explanation: