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madam [21]
2 years ago
13

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12

. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .05 significance level, what is the p-value for testing variances for a one-tail test?
Mathematics
1 answer:
marissa [1.9K]2 years ago
7 0

Answer:

Step-by-step explanation:

Hello!

You have two random samples obtained from two different normal populations.

Sample 1

n₁= 15

X[bar]₁= 350

S₁= 12

Sample 2

n₂= 17

X[bar]₂= 342

S₂= 15

At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.

If the statistic hypotheses are:

H₀: σ₁² ≥ σ₂²

H₁: σ₁² < σ₂²

The statistic to test the variances ratio is the Stenecor's-F test. F_{H_0}=(\frac{S^2_1}{Sigma^2_1}) * (\frac{S^2_2}{Sigma^2_2} )~F_{n_1-1;n_2-1}

F_{H_0}= \frac{(12)^2}{(15)^2} * 1= 0.64

The p-value is:

P(F_{14;16}≤0.64)= 0.02

I hope it helps!

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Assume that the die is weighted so that the probability of a 1 is 0.1, the probability of a 2 is 0.2, the probability of a 3 is
umka2103 [35]

Answer:

The probabilities of each outcome are the following:

for X = 1 is 0.6

for X = 2 is 0.35

for X = 3 is 0.049

and for X = 4 is 0.001

Step-by-step explanation:

Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is: S\geq 4

Let's now analyze and consider the ways that X equals the different values:

X = 1: I throw the dice and the result is: 4, 5 and 6.

Then: P(X=1) = P(4) + P(5) + P(6) = 0.2+0.1+0.3 = 0.6

Thus P(X=1) = 0.6

X = 2: The result after following the dice two times can be:

1 and 3, 1 an 4.... and so on until 1 and 6

2 and 2, 2 and 3... and so on until 2 and 6

3 and 1... and so on until 3 a 6

Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)

Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1

Then P(1......6) = 1

Then P(X=2) = P(1)x[1-P(1)-P(2)]+P(2)x[1-P(1)]+P(3) = 0.1x(1-0.1-0.2) + 0.2x(1-0.1) + 0.1 = 0.1 x 0.7 + 0.2 x 0.9 + 0.1 = 0.35

Thus P(X=2) = 0.35

X = 3: The result can be

1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6

1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6

2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6

Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)

P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049

Thus P(X=3) = 0.049

Finally, for X to be 4, I only have the following possibilities

1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6

Then P(X=4) = P(1)xP(1)xP(1)xP(1.....6) = 0.1x0.1x0.1x1 = 0.001

Thus P(X=3) = 0.001

5 0
2 years ago
a city us growing at the rate of 0.9% annually. if there were 3619000 residents in yhe city in 1994, find how many to the neares
BlackZzzverrR [31]
For this case we have the following equation:
 y = 3619000 (2.7) ^ 0.009t
 We must evaluate the equation for the year 2000.
 Therefore, we must replace the following value of t:
 t = 2000 - 1994
 t = 6
 Substituting we have:
 y = 3619000 (2.7) ^ (0.009 * 6)
 y = 3818407.078
 Round to the nearest ten thousand:
 y = 3820000
 Answer:
 
3820000 residents are living in that city in 2000
8 0
2 years ago
Verify the continuity type C° and C1 between curve(l) and curve(2).
Damm [24]

Step-by-step explanation:

to be honest I'm not sure how to do

Comment

6 0
2 years ago
A carpool service has 2,000 daily riders. A one-way ticket costs $5.00. The service estimates that for each $1.00 increase to th
solmaris [256]

Answer:

Total number of riders that ride on carpool daily = 2000

Total Cost of one way ticket = $ 5.00

Total Amount earned if 2000 passengers rides daily on carpool = 2000 × 5

                                                                                                            = $10,000

If fare increases by $ 1.00

New fare = $5 + $1    

               = $6

Number of passengers riding on carpool = 2,000 - 100 = 1,900

If 1,900 passengers rides on carpool daily , total amount earned ,if cost of each ticket is $ 6 = 1900 × $6 = $11400

As we have to find the inequality which represents the values of x that would allow the carpool service to have revenue of at least $12,000.

For $ 1 increase in fare = (2,000 - 1 × 100) passengers

For $ x increase in fare, number of passengers = 2,000 - 100·x

                                                         = (2,000 - 100·x) passengers

New fare = 5 + x

New Fare × Final Number of passengers ≥ 12,000

(5+x)·(2,000 - 100 x) ≥ 12,000

5 (2,000 - 100 x) + x(2,000 - 100 x) ≥ 12,000

10,000 - 500 x + 2,000 x - 100 x² ≥ 12,000

100 - 5 x + 20 x - x² ≥ 120

- x² + 15 x +100 - 120 ≥ 0

-x² + 15 x -20 ≥ 0

x² - 15 x + 20 ≤ 0

⇒ x = 1.495

x ≥ $ 1.495, that is if we increase the fare by this amount or more than this the revenue will be at least 12,000 or more .

Also, f'(x) = 0 gives x = 7.5

⇒ The price of a one-way ticket that will maximize revenue is $7.50

7 0
2 years ago
Kara starts tying a baby quilt at the local community center. Each quilt takes 200 ties to complete. Ten minutes later Julie joi
Tju [1.3M]

Answer:

a) K(t)=6\frac{tie}{min}(t)

b)J(t)=11\frac{tie}{min}(t)

c)T_{f}(t)=K(t+10)+J(t) where T_{f}(t)= number ties finished per unit time

d) t≅8.24min

e) K=109tie

   J=91tie

f) K≅16.6min

  J≅9.09min

Step-by-step explanation:

Data

Kara_{(K)}=6\frac{tie}{min}\\Julie(J)=11\frac{tie}{min}

Kara begins ten minutes before julie, so when julie joins kara she has 60 ties

t=10 k=6\frac{tie}{min}*10min=60tie

a) K(t)=6\frac{tie}{min}(t)

b)J(t)=11\frac{tie}{min}(t)

c)T_{f}(t)=K(t+10)+J(t) where T_{f}(t)= number ties finished per unit time

d) if kara already has 60 ties and makes 6 per minute and Julie makes 11 per minute then per minute they makes 17 ties, so one minute per 17 ties for 200 ties ¿how long?. t=\frac{(200-60)tie*min}{17tie}≅8.24min

e) K=60tie+6tie/min*8.24min=109tie

   J=11tie/min*8.24min=91tie

f) K=100tie*min/6tie≅16.6min

  J=100tie*min/11tie≅9.09min

Note: the amount of ties were rounded to the nearest decimal

6 0
2 years ago
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