Answer two questions about Systems A and B:System A x+3y=−9 2x+y=4 System B 3x+4y=−9 2x+y=4

A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.

krek1111 [17]

Answer:

a) $\hat p=\frac{1627}{5120}=0.3178$

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

$X=1627$ number of permanent dwellings on the entire reservation that are traditional hogans.

$n=5120$ random sample taken

$\hat p=\frac{1627}{5120}=0.318$ estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

$p$ true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

$\hat p=\frac{1627}{5120}=0.3178$

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.318 - 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.301$

$0.318 + 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.335$

The 99% confidence interval would be given by (0.301;0.355)

8 0

2 years ago

What is the solution to -13×=78

Pavlova-9 [17]

-13x=78
-13x/-13=78/-13
Answers: X=-6

8 0

2 years ago

Read 2 more answers

Complete the following tables by filling in the missing values and determine the rule for each table:

Nesterboy [21]

Answer:

I really don't know just answering to get points

5 0

1 year ago

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Gradien dari persamaam 3x + y = 7 adalah

Veronika [31]

Answer:

Step-by-step explanation:

6 0

2 years ago

Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t

STatiana [176]

Answer:

Part a) Probability that a moose in that age group is killed by a wolf

P (0.5 year) = 0.361

P (1 - 5) = 0.159

P (6 - 10) = 0.267

P (11 - 15) = 0.203

P (16 - 20) = 0.010

Part b)

Expected age of a moose killed by a wold

μ = 5.61 years

Stantard deviation of the ages

σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

Age of Moose in years Number killed by woolves

Calf (0.5) 107

1 - 5 47

6 - 10 79

11 - 15 60

16 - 20 3

You can now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) First question: a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) Second question: b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class midpoint

0.5 0.5

1 - 5 (1 + 5) / 2 = 3

6 - 10 (6 + 10) / 2 = 8

11 - 15 (11 + 15) / 2 = 13

16 - 20 (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s = (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8 - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

7 0

2 years ago