Recent questions in Statistics and Probability

Samples
Answered

Kaila Branch
2022-09-27

$a11\phantom{\rule{0ex}{0ex}}b.10\phantom{\rule{0ex}{0ex}}c.14\phantom{\rule{0ex}{0ex}}d.8$

Fisher's Exact Test
Answered

misyjny76
2022-09-26

Survey Questions
Answered

Dymnembalmese2n
2022-09-26

I have solved this question but I don't know whether the answer is correct or not.

H0: mean = 120 (null hypothesis) H1: mean > 100 (alternative hypothesis)

we will use z test as the sample count is more than 30

$z=|120-100|/60/\sqrt{64}\phantom{\rule{0ex}{0ex}}z=2.67$

at 5% of significance, the critical value of z is 1.96. Since the z value we obtained is more than 1.96, so we reject the null hypothesis and therefore the mean price of the test is more than 100

Please tell whether the answer is correct or there is some mistake in this. Help is appreciated.

T-Statistic
Answered

unjulpild9b
2022-09-26

Survey Questions
Answered

Nathanael Perkins
2022-09-26

A survey organisation wants to take a simple random sample in order to estimate the percentage of people who have seen a certain programme. The sample is to be as small as possible. The estimate is specified to be within 1 percentage point of the true value; i.e., the width of the interval centered on the sample proportion who watched the programme should be 1%. The population from which the sample is to be taken is very large. Past experience suggests the population percentage to be in the range 20% to 40%. What size sample should be taken?

I think I have to use this and solve for n

$$1.96\sqrt{\frac{\pi (1-\pi )}{n}}=.01$$

where $\pi $ is the sample estimated proportion of people who watch the programme.

Now does this mean that I am 95% sure that I am within 1% accuracy? I also am not aware as to how I can find π though I have read I could use the population standard deviation instead and suspect I would have to use that as I am given some information- that the pop proportion is 20%-40%.

Finally in general what is being said here:

$$\pi \pm 1.96\sqrt{\frac{\pi (1-\pi )}{n}}=.01$$

My notes at the moment just say it contains the population mean 95% of the time.... why? I think if I had some graphical understanding of what was going on everything would be much simpler for me.

Pooled Standard Deviation
Answered

fion74185296322
2022-09-26

Sampling Frame
Answered

imchasou
2022-09-26

Survey Questions
Answered

Haven Kerr
2022-09-25

So here's the problem:

The table below shows the distribution of education level attained by US residents based on data collected during the 2010 American Community Survey:

Highest level of education %

Less than 9th Grade 0.10

9th to 12th no diploma 0.09

High school grad - GED 0.25

Some college No degree 0.23

Associate's degree 0.08

Bachelor's degree --

Graduate or professional degree 0.09

Answer the following questions (give all answers to 2 decimal places):

a) Fill in the empty box for the proportion of US residents whose highest education level attained was a bachelors degree.

b) If two individuals are chosen at random from the population, what is the probability that both will have at least a bachelors degree?

c) If two individuals are chosen at random from the population, what is the probability that at least one will have some college or a college degree of some sort?

d) If two individuals are chosen at random from the population, what is the probability that exactly one will have some college or a college degree of some sort?

I managed to figure out a) which was 0.16. However, I tried a different methods to get b) but none of them worked. I answered .24, .25, and .01 but none of them were correct.

To answer b) I used the following formula:

P(A or B) = P(A) + P(B) - P(A & B) = (.16) + (.09) - (.16)(.09)

which gave me .24 but that was incorrect. What am I doing wrong?

Random variables
Answered

Quinlan7g
2022-09-25

Random variables
Answered

Medenovgj
2022-09-25

Survey Questions
Answered

Kallie Fritz
2022-09-25

The reason I am asking is that I vaguely remember there are statistical sampling techniques from college days that survey companies used to represent the bigger populations. For instance, after surveying 2000 people, they can conclude something for the U.S. population. In other words, they "know" 2000 people can represent the U.S. population with some scientific backing.

Just curious, if there is a way to figure out the magical number "x", then I can go with the lowest rate after browsing through "x" vendors as well. The question is purely theoretical in this simple case: the lowest rate wins. Any leads or ideas are appreciated.

Random variables
Answered

memLosycecyjz
2022-09-25

Confidence intervals
Answered

Camila Brandt
2022-09-25

Suppose we use a sample mean $\overline{X}$ to construct a 95% confidence interval [a,b].

I was told that it is incorrect to say that there is a 95% probability that the population mean lies between a and b. Because the population mean is a constant and not a random variable. The probability that a constant falls within any given range is either 0 or 1.

However, from the textbook it is said that we expect 95% of the confidence intervals to include the population mean.

If 95% of the confidence intervals are expected to include the population mean, then each confidence interval has 95% probability to include the population mean. Therefore I think it is correct to say there is a 95% probability that the population mean lies between a and b.

Where did I make mistakes?

Comparing two groups
Answered

Alexus Deleon
2022-09-25

A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.

Solution: Creating an indicator r.v. for each pair of people, we have that the expected number of pairs of people with the same birthday is (50C2 . 1/365) by linearity. Now create an indicator r.v. for each day of the year, taking the value 1 if at least two of the people were born that day (and 0 otherwise). Then the expected number of days on which at least two people were born is

$365(1-(364/365{)}^{50}-50\cdot (1/365)\cdot (364/365{)}^{49})$

Could someone explain how we got the answer ?

T-Statistic
Answered

gaby131o
2022-09-25

$t=\frac{\overline{x}-\mu}{s}$

How is the sample estimate of the standard error computed in this context? In particular, does it make use of the population mean or the sample mean?

Population Standard Deviation
Answered

videosfapaturqz
2022-09-25

Cluster Sampling
Answered

Altenbraknz
2022-09-25

1. To get a sense of election outcomes, a political group chooses ten precincts to conduct a survey of voters in those areas.

Confidence intervals
Answered

waldo7852p
2022-09-25

However, to my knowledge, isn't this one of the common misconceptions of a confidence interval? And that the actual interpretation is that if we gather n amount of these confidence intervals, there is a 95% probability that the n collected intervals all contain the true parameter?

My notes give this definition:Let $L:=L({X}_{1},\dots ,{X}_{n})$ and $U:=U({X}_{1},\dots ,{X}_{n})$ be such that for all $\theta \in \mathrm{\Theta}$,

$$\mathbb{P}(L<\theta \le U)\ge 1-\alpha $$

and then it says, that this is the probability $1-\alpha $ that the true parameter lies within this interval . Which one is correct?

Random variables
Answered

lunja55
2022-09-25

Confidence intervals
Answered

Alexus Deleon
2022-09-25

For this question from a large amount of data I have calculated that the mean is 44.22, the sample size is 100 and the standard deviation is 22.0773.

From this I am asked to , make the 98% confidence intervals for the (1) true mean µ of the module mark (2) true variance of the module mark

And for each: (a) Determine what quantity to look at, and which distribution table to use, justifying your choice. (b) Determine the number of degrees of freedom, justifying your answer. (c) Calculate the actual intervals.

So far for 1, I have used the z table to look for 99% as I need 1% to the right of 2.33 and 1% to the left of −2.33, so 98% is between $\pm 2.33$. Giving me

$$\overline{x}\pm 2.33\frac{\sigma}{\sqrt{n}}$$

Which provides me with a 39.08 to 49.36 confidence interval, is this correct? And how would I determine degrees of freedom and go about answering part 2?

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