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Artist 52 [7]
3 years ago
15

Choose the correct definition of a sampling distribution. The sampling distribution of a statistic of size n is:_______a. The di

stribution of all values of the statistic resulting from all samples of size n n taken from the same population. b. The distribution of the statistic in one simple random sample of size n n from a given population. c. The distribution of the parameter in all possible populations of size n n. d. The distribution of the values obtained from a simple random sample of size n n from the same population. e. The distribution of the parameter in one simple random sample of size n n from a given population.
Mathematics
1 answer:
Grace [21]3 years ago
7 0

Answer:

The correct option is d)

The distribution of the values obtained from a simple random sample of size n from the same population

Step-by-step explanation:

Sampling distribution is the process of getting sample through simple random techniques from the population.

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Answer:

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Step-by-step explanation:

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3a^2 (ab^2+b^2)=<br> How much
kiruha [24]

Answer is 3a²b (a + b)

Step-by-step explanation:

  • Step 1: Find 3a² (ab²+b²)

⇒ 3a² (ab²+b²) = 3a³b + 3a²b² = 3a²b (a + b)

8 0
3 years ago
-52 times the difference between a number and 14 is equal to the number plus 3
xenn [34]
X=number looked for
we have this equation:
-52(x-14)=x+3
we solve this equation:
-52x+728=x+3
-52x-x=3-728
-53x=-725
x=-725/-53=725/53 ≈13,68

Solution: 725/53

To check:
the difference between 725/53 and 14=725/53  - 14=(725-742)/53=-17/53
-52 times the diference is= -52(-17/53)=884/53
The number plus 3=725/53 + 3=(725+159)/53=884/53
3 0
3 years ago
There are 500 toothpicks in a small box. If a jumbo box is made by doubling all the dimensions of the small box, how many toothp
Kisachek [45]
Volume is length x width x height. if you double those it is 2length x 2width x 2 height so you are increasing the volume by 8 times (2x2x2). 8 x 500 = 4000<span>                                   D.) 4,000 Toothpicks
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4 0
3 years ago
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A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o
ryzh [129]

Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is  p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑_{x=0 b( x, n, p )

= ³∑_{x=0 b( x, 10, 0.6 )

= ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.6)^x( 1 - 0.6 )^{10-x

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.3 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.4 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.5 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

3 0
2 years ago
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