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r-ruslan [8.4K]
2 years ago
11

A sportswriter wishes to see if a football filled with helium travels farther, on average, than a football filled with air. To t

est this hypothesis, the writer uses 18 male subjects, randomly divided into two groups of 9 subjects each. Group 1 kicks a football filled with helium to the recommended pressure, while Group 2 kicks a football filled with air to the same pressure. The sample mean yardage for Group 1 was barx2 = 30 yards with a standard deviation of s1 = 8 yards. The mean yardage for Group 2 was barx2 = 25 yards with a standard deviation of s2 = 6 yards. Assume that the two groups of kicks are independent. Let μ1 and μ2 represent the mean yardage we would observe for the entire population represented by the volunteers if all members of this population kicked, respectively, a helium-filled football and an air-filled football. Let σ1 and σ2 be the corresponding population standard deviations. Assuming two-sample t procedures are safe to use, what is a 99% confidence interval for μ1 – μ2?
A) 4 ± 7.7 yards.B) 4 ± 11.2 yards.C) 4 ± 4.7 yards.D) 4 ± 6.2 yards.
Mathematics
1 answer:
Sphinxa [80]2 years ago
4 0

Answer:

5 ± 9.74

Step-by-step explanation:

Confidence interval, two - sample, t procedure ;

μ1 - μ2 = (x1 - x2) ± Tcritical*S

S = sqrt[(s1²/n1) + (s2²/n2)]

n1 = 9 ; n2 = 9 ; s1 = 8 ; s2 = 6 ; x1 = 30 ; x2 = 25

S = sqrt[(8^2/9) + (6^2/9)]

S = sqrt[11.111111]

S = 3.333

Tcritical at 99% confidence interval

df = 18 - 2= 16

Tcritical = 2.92

(30 - 25) ± 2.92*(3.333)

5 ± 9.74

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