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3 years ago

When a higher confidence level is used to estimate a proportion and all other factors involved are held constant

Mathematics

1 answer:

lbvjy [14]3 years ago

8 0

Options :

A) the confidence interval will be wider.

B) the confidence interval will be narrower.

C) there is not enough information to determine the effect on the confidence interval.

D) the confidence interval will be less likely to contain the parameter being estimated.

E) the confidence interval will not be affected.

Answer: A) The confidence interval will be wider

Step-by-step explanation: In statistics, When a higher confidence interval is used to estimate a proportion with other factors being held constant, the confidence interval will be wider. Since the confidence interval is aimed at tweaking the precision of our measurement or estimation, when we adopt a higher confidence level, we are trying g to reduce our chances of being wrong and hence increasing our precision and thus widening the level of confidence in the result obtained. That said, 95% confidence interval is wider than 90%. And 99% is wder than 95%.

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What is the answer to 6-1x0+2/2 ?

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3 years ago

A box of tickets has an average of 420; the SD is 84. If we draw at random (with replacement) 50 times and compute the average o

disa [49]

Answer:

Average of the draws equals 420

Standard Error = 11.88

Step-by-step explanation:

Given

$\mu = 420$

$\sigma = 84$

$n =50$

Solving (a): The average of the draws

This implies that we calculate the sample mean

This is calculated as:

$\bar x = \mu$ --- Sample Mean = Population Mean

So, we have:

$\bar x = 420$

Solving (b): The standard error

This is calculated as:

$SE=\frac{\sigma}{\sqrt n}$

So, we have:

$SE=\frac{84}{\sqrt {50}}$

Using the calculator, we have:

$SE=11.88$

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3 years ago

Slips of paper numbered 1 to 15 are placed in a box. A slip of paper is drawn at random. What is the probability that the number

monitta

Given:

Slips of paper numbered 1 to 15 are placed in a box.

To find:

The probability that the number picked is either a multiple of 5 or an odd number.

Solution:

We have,

Total outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

No. of total outcomes = 15

Multiple of 5 are 5, 10, 15.

Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15.

Number that are either a multiple of 5 or an odd number are 1, 3, 5, 7, 9, 10, 11, 13, 15.

No. of favorable outcomes = 9

We know that,

$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$\text{Probability}=\dfrac{9}{15}$

$\text{Probability}=\dfrac{3}{5}$

$\text{Probability}=0.6$

Therefore, the probability that the number picked is either a multiple of 5 or an odd number is 0.6.

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3 years ago

If you could get the answer with a short explanation that would be great!

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Answer:

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3 years ago

Read 2 more answers

Marianna [84]

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Step-by-step explanation:

Let us revise some logarithmic rules

$log(a)^{n}=nlog(a)$
$log(ab)=log(a)+log(b)$
$nlog(a)+mlog(b)=log[(a)^{n}][(b)^{m}]$

∵ 10 log(x) + 5 log(64)

- At first re-write 10 log(x)

∴ 10 log(x) = $log(x)^{10}$

- Then re-write 5 log(64)

∴ 5 log(64) = $log(64)^{5}$

∴ 10 log(x) + 5 log(64) = $log(x)^{10}$ + $log(64)^{5}$

- Use the 3rd rule above to make it single logarithm

∵ $log(x)^{10}$ + $log(64)^{5}$ = $log[(x)^{10}][(64)^{5}]$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(64)^{5}]$

∵ 64 = 2 × 2 × 2 × 2 × 2 × 2

∴ We can write 64 as $2^{6}$

∴ $(64)^{5}=(2^{6})^{5}$

- Multiply the two powers of 2

∴ $(64)^{5}=(2)^{30}$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(2)^{30}]$

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

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3 years ago