what happens to the width of the confidence interval as the confidence level increases?

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Which of the post-obit statements are correct about confidence intervals?

Possible Answers:

The width of a confidence interval decreases equally 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 as the confidence 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 confidence interval does not change as the sample size increases and increases equally the confidence level increases.

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

Correct respond:

The width of a conviction interval decreases every bit the sample size increases and increases every bit 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 confidence level increases the take chances that the correct value will be found in the confidence interval. This means that the interval is larger.

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

Correct answer:

Explanation:

Keep in mind that the margin of fault 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 problem, nosotros tin tell that $z^{*} = 1.96$ and $\sigma = 5$ . We can 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 upwards, which is . If you are unsure on issues like these, you tin can bank check the margin of error for your answer rounded down and then rounded upward (in this case, for n = 10 and n = 11 .)

Jim calculated a $95\%$ confidence interval for the mean top in inches of boys in his loftier schoolhouse. He is not certain how to interpret this interval. Which of the following explains the meaning of $95\%$ conviction.

Possible Answers:

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

There is a $95\%$ chance that Jim's interval contains the truthful hateful acme.

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

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

More data is needed.

Correct answer:

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

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 hateful that in that location is a 95% chance that the true mean will exist within the interval. It also does not mean that 95% of all heights or possible sample means autumn within the interval.

The number of hamburgers served by McGregors per solar day is normally distributed and has a mean of 3,250 hamburgers and a standard deviation of 320. Discover the range of customers served on the middle percent of days.

Right answer:

(3,034, 3,466)

Explanation:

Beginning, discover the first quartile of the distribution.

$P(X < Q_{1}) = 0.25 \Rightarrow Q_{1} = \mu - Z_{0.25}\sigma = 3,250 - 0.675(320) = 3,034$

Then, discover the tertiary quartile of the distribution.

$P(10 < Q_{3}) = 0.75 \Rightarrow Q_{3} = \mu + Z_{0.25}\sigma = 3,250 + 0.675(320) = 3,466$

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

Correct answer:

0.65

Explanation:

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

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

Assume in that location is an election involving 3 parties: D, R, and I. The probability of D winning is .eleven, R winning is .78, and I winning is .xi. What is the probability of D or R winning?

Explanation:

Since all of the events are mutually exclusive (1 of the parties must win), you can get the probability of either D or R winning by adding their probabilities.

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

If one bill of fare is chosen at random from a deck of cards, what is the probability that information technology will be a heart or a king?

Possible Answers:

4/thirteen

13/52

26/52

17/52

29/52

Explanation:

In a deck of cards, in that location are 52 full cards, thirteen hearts, iv kings, and ane rex that is a eye.

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

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

Right answer:

$\frac{3}{13}$

A educatee randomly selected a highlighter from her desk. There were five highlighters on the desk-bound, each of a unlike color--bluish, green, yellow, red, and orange. What is the probability that the student selected either the carmine or the yellow highlighter?

Correct respond:

2/5

Explanation:

In this case, we desire to know the probability of multiple, mutually exclusive possible outcomes. To determine the probability of the two possible outcomes, but add them together. This is called the add-on rule.

1/5+1/5=2/5

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

Correct reply:

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

Caption:

First find the number of ways to go 0, 1, or 2 heads.

Call back, $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ .Likewise, $\binom{n}{0}=1$ and $\binom{n}{1}=n$ .

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

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

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 two heads in x money 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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what happens to the width of the confidence interval as the confidence level increases?

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