a)

b)

c)

Rivka Thorpe
2021-02-10
Answered

Suppose you have selected a random sample of ? $=14$ measurements from a normal distribution. Compare the standard normal ? values with the corresponding ? values if you were forming the following confidence intervals.

a)$95\mathrm{\%}$ confidence interval

$?=$

$?=$

b)$80\mathrm{\%}$ confidence interval

$?=$

$?=$

c)$98\mathrm{\%}$ confidence interval

$?=$

$?=$

a)

b)

c)

You can still ask an expert for help

lobeflepnoumni

Answered 2021-02-11
Author has **99** answers

Step 1

Given:

A random sample of 14 measurements from a normal distribution is selected. That is,

Sample size:$n=14$

Let, X be the measurement values.

If some data {x} are normally distributed, the corresponding {z} will be normal with mean 0 and standard deviation 1 where the correspondence between the {x} and {z} is given by$z=\frac{x-\mu}{\sigma}$ . Such zs

Given:

A random sample of 14 measurements from a normal distribution is selected. That is,

Sample size:

Let, X be the measurement values.

If some data {x} are normally distributed, the corresponding {z} will be normal with mean 0 and standard deviation 1 where the correspondence between the {x} and {z} is given by

asked 2021-03-09

In a study of the accuracy of fast food drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part (a) to this 90% confidence interval for the percentage of orders that are not accurate at Restaurant B:

asked 2021-08-04

A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.

Assume the distribution of measurements to be approximately normal.

a) Construct a$99\mathrm{\%}$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.

b) What can we assert with$99\mathrm{\%}$ confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

Assume the distribution of measurements to be approximately normal.

a) Construct a

b) What can we assert with

asked 2020-12-27

Consider the next 1000 98% Cis for mu that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of $\mu ?$

What isthe probability that between 970 and 990 of these intervals conta the corresponding value of ? (Hint: Let

Round your answer to four decimal places.)

‘the number among the 1000 intervals that contain What king of random variable s 2) (Use the normal approximation to the binomial distribution

What isthe probability that between 970 and 990 of these intervals conta the corresponding value of ? (Hint: Let

Round your answer to four decimal places.)

‘the number among the 1000 intervals that contain What king of random variable s 2) (Use the normal approximation to the binomial distribution

asked 2021-02-23

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let j: denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence intervals (7-8, 9.6)

(a) Would 2 90%% confidence interval calculated from this same sample have been narrower or wider than the glven interval? Explain your reasoning.

(b) Consider the following statement: There is 9 95% chance that Is between 7.8 and 9.6. Is this statement correct? Why or why not?

(c) Consider the following statement: We can be highly confident that 95% of al bottles ofthis type of cough syrup have an alcohol content that is between 7.8 and 9.6. Is this statement correct? Why or why not?

(a) Would 2 90%% confidence interval calculated from this same sample have been narrower or wider than the glven interval? Explain your reasoning.

(b) Consider the following statement: There is 9 95% chance that Is between 7.8 and 9.6. Is this statement correct? Why or why not?

(c) Consider the following statement: We can be highly confident that 95% of al bottles ofthis type of cough syrup have an alcohol content that is between 7.8 and 9.6. Is this statement correct? Why or why not?

asked 2021-02-05

Give a correct answer for the difference between internal and external validity is best described as:

asked 2022-04-16

Weaknesses of Wald confidence interval for binomial distribution

A survey samples 1000 people, among which 500 say they will vote for A, 400 for B e 100 for C. Calculate a confidence interval for the proportion of people that will vote for A. What are the major weaknesses of the confidence interval you just calculated?

Let${A}_{n}\sim \text{Binom}(n,p)$ . By the CLT,

$\sqrt{n}({\stackrel{\u2015}{A}}_{n}-p)x\to \left\{d\right\}\mathcal{N}(0,p(1-p))$

hence$\frac{\sqrt{n}}{\sqrt{p(1-p)}}({\stackrel{\u2015}{A}}_{n}-p)x\to \left\{d\right\}\mathcal{N}(0,1)$

Since p is unknown, we approximate it with$\stackrel{\u2015}{A}}_{n$ . It follows that, at $1-\alpha$ confidence,

$p\in {\stackrel{\u2015}{A}}_{n}\pm {z}_{1-\frac{\alpha}{2}}\sqrt{\frac{{\stackrel{\u2015}{A}}_{n}(1-{\stackrel{\u2015}{A}}_{n})}{n}}$

Since this confidence interval relies on the CLT, it gives poor results when n is small or p is very close to 0 or 1. But this is not the case here, as$n=1000$ and ${\stackrel{\u2015}{A}}_{n}=0.5$ , so I fail to see what weaknesses the question is talking about?

A survey samples 1000 people, among which 500 say they will vote for A, 400 for B e 100 for C. Calculate a confidence interval for the proportion of people that will vote for A. What are the major weaknesses of the confidence interval you just calculated?

Let

hence

Since p is unknown, we approximate it with

Since this confidence interval relies on the CLT, it gives poor results when n is small or p is very close to 0 or 1. But this is not the case here, as

asked 2022-04-22

How can I find the Margin of Error E

The professor hasn't gone over this section but I want to understand it so that I won't confused in class. I want to learn so that I can do the other problems.

So this is the problem, If anyone could teach me so I could take note I'd deeply appreciate it. Thank you for your time.

Question: Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.$n=550$ , x equals 330, 90 % confidence

I know how to obtain the alpha/2 to find the confidence.

I just went over the Binomial Distribution and Limit Theorem, which understand now this is next. If anyone would care to spent their time explaining I'd be forever grateful, thank you again for your time.

The professor hasn't gone over this section but I want to understand it so that I won't confused in class. I want to learn so that I can do the other problems.

So this is the problem, If anyone could teach me so I could take note I'd deeply appreciate it. Thank you for your time.

Question: Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.

I know how to obtain the alpha/2 to find the confidence.

I just went over the Binomial Distribution and Limit Theorem, which understand now this is next. If anyone would care to spent their time explaining I'd be forever grateful, thank you again for your time.