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avanturin [10]
3 years ago
11

Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140

graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute.
a. To compare the pass rates, what would be the pooled proportion?

b. What is the test statistic?

c. Find the critical value for a right-tailed test at ? = .05.

d. Find the p-value.

d. What is your decision regarding the null hypothesis?
Mathematics
1 answer:
egoroff_w [7]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that two well-known aviation training schools are being compared using random samples of their graduates

Fly more academy 70 of 140

Blue Yonder   104 of 260

Combined pass = (70+104) out of (140+260)

a) Pooled proportion=\frac{174}{400} \\=0.435

b) H0: p1 = p2

Ha: p1 ≠p2

(two tailed test)

p difference= \frac{70}{140} -\frac{104}{260} =0.10

std error for difference (using pooled proportion) = \sqrt{\frac{0.435*0.565}{400} } \\=0.0248

Test statistic = p difff/std error = 4.034

c) Critical value for 0.05 is 1.96

d) p value is < 0.005

Since p < 0.05 our significant level we reject H0

There is significant difference between the two proportions.

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The altitude of a triangle is increasing at a rate of 1 cm/ min while the area of the triangle is increasing at a rate of 2 cm2
serg [7]
Ya, calculus and related rates, such fun!




everything is changing with respect to t

altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min

area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min

base=db/dt

alright

area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt

ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b

so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min

therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt

the base is decreasing at 1.6cm/min
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