if confidence level increases what happens to width

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Which of the following statements are correct nearly confidence intervals?

Possible Answers:

The width of a confidence interval decreases every bit the sample size increases and increases as the confidence level decreases.

The width of a confidence interval increases as the sample size increases and increases equally the conviction level decreases.

The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.

The width of a conviction interval does not modify every bit the sample size increases and increases as the confidence level increases.

The width of a conviction interval increases as the sample size increases and increases equally the confidence level increases.

Correct reply:

The width of a confidence interval decreases equally the sample size increases and increases as the confidence level increases.

Explanation:

Larger samples give narrower intervals. We are able to estimate a population proportion more precisely with a larger sample size.

As the confidence level increases the width of the conviction interval also increases. A larger conviction level increases the risk that the correct value will be found in the conviction interval. This means that the interval is larger.

You are asked to create a $95\%$ conviction interval with a margin of fault no larger than while sampling from a commonly distributed population with a standard divergence of . What is the minimum required sample size?

Right respond:

Explanation:

Go along in mind that the margin of error for a confidence interval based on a normal population is equal to $z^{*}\left(\frac{\sigma}{\sqrt{n}}\right)$ , where $z^{*}$ is the -score corresponding to the desired confidence level.

From the trouble, we tin can tell that $z^{*} = 1.96$ and $\sigma = 5$ . Nosotros can and then solve for algebraically:

$3 = 1.96\left(\frac{5}{\sqrt{n}}\right)$

$\frac{3}{1.96} = \frac{5}{\sqrt{n}}$

$\frac{3}{1.96}\cdot \frac{1}{5} = \frac{1}{\sqrt{n}}$

$n =\left ( \frac{3}{1.96 \cdot 5} \right )^{-2} = 10.671$

The minimum sample size is 10.671 rounded up, which is . If you are unsure on problems like these, you tin check the margin of error for your answer rounded down and and then rounded upwards (in this case, for n = 10 and n = 11 .)

Jim calculated a $95\%$ confidence interval for the mean acme in inches of boys in his high schoolhouse. He is non certain how to interpret this interval. Which of the following explains the meaning of $95\%$ confidence.

Possible Answers:

In the long run, $95\%$ of all confidence intervals calculated from the same population will incorporate the true mean height.

There is a $95\%$ take a chance that Jim'due south interval contains the true mean acme.

$95\%$ of boys' heights fall with the interval Jim calculated.

$95\%$ of all possible sample means fall within Jim's interval.

More than data is needed.

Correct answer:

In the long run, $95\%$ of all confidence intervals calculated from the same population volition contain the true mean summit.

Explanation:

95% confidence means that the methods Jim uses to summate his confidence interval requite him correct results 95% of the time. It does not mean that at that place is a 95% chance that the true hateful will exist inside the interval. It as well does not mean that 95% of all heights or possible sample means autumn within the interval.

The number of hamburgers served by McGregors per twenty-four hours is unremarkably distributed and has a mean of 3,250 hamburgers and a standard deviation of 320. Detect the range of customers served on the middle percent of days.

Correct answer:

(3,034, 3,466)

Caption:

First, find the get-go quartile of the distribution.

Then, find the tertiary quartile of the distribution.

The probability that it will rain today is 0.35. What is the probability that information technology will not pelting?

Correct reply:

0.65

Caption:

The reply is 0.65 because Pr(~Rain) is the complement of Pr(Pelting) and both events are mutually exclusive.

When two events are mutually sectional, $Pr(A\ and\ B)=Pr(A)+Pr(B)$ . Since probabilities must sum up to 1, this implies that Pr(B)=1-Pr(A) .

Assume in that location is an ballot involving three parties: D, R, and I. The probability of D winning is .11, R winning is .78, and I winning is .11. What is the probability of D or R winning?

Caption:

Since all of the events are mutually exclusive (one of the parties must win), you can become the probability of either D or R winning by calculation their probabilities.

Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89.

If 1 bill of fare is chosen at random from a deck of cards, what is the probability that information technology volition be a middle or a king?

Possible Answers:

4/13

13/52

26/52

17/52

29/52

Explanation:

In a deck of cards, there are 52 full cards, 13 hearts, 4 kings, and i king that is a middle.

And so, 13/52 + 4/52 - 1/52 = 16/52 = 4/13.

Using a standard deck of cards, what is the probability of choosing a unmarried face card?

Correct respond:

$\frac{3}{13}$

A pupil randomly selected a highlighter from her desk. At that place were five highlighters on the desk, each of a different color--blue, green, yellow, red, and orange. What is the probability that the student selected either the red or the yellow highlighter?

Right reply:

2/5

Explanation:

In this case, nosotros want to know the probability of multiple, mutually exclusive possible outcomes. To make up one's mind the probability of the two possible outcomes, just add together them together. This is called the addition rule.

1/5+1/5=2/5

What is the probability of obtaining at most heads when tossing a fair coin times?

Correct reply:

$p = \frac{56}{1024}$

Explanation:

Showtime detect the number of ways to become 0, 1, or 2 heads.

Think, $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ .Also, $\binom{n}{0}=1$ and $\binom{n}{1}=n$ .

heads: $\binom{10}{0} = 1$

head: $\binom{10}{1} = 10$

heads: $\binom{10}{2} = 45$

$\textup{number of possible outcomes of 10 tosses}= 2^{10} = 1024$

$p\left ( x\right )=\frac{\textup{number\ of\ desired\ outcomes}}{\textup{number\ of\ all\ possible\ outcomes}}$

$p(\textup{0 heads in 10 coin tosses}) = \frac{1}{1024}$

$p(\textup{1 head in 10 coin tosses}) = \frac{10}{1024}$

$p(\textup{2 heads in 10 coin tosses}) = \frac{45}{1024}$

Now combine these to find the probability of seeing at most 2 heads in x coin tosses:

$p(\textup{at most 2 heads in 10 coin tosses})$

$=p(\textup{0 heads})+p(\textup{1 head})+p(\textup{2 heads})$

$=\frac{1}{1024} +\frac{10}{1024} +\frac{45}{1024}$

$=\frac{56}{1024}$

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if confidence level increases what happens to width

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