Recent questions in Research Methodology

Average
Answered

Ruby Briggs
2022-07-23

Nonresponse Bias
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pliwraih
2022-07-23

1. Nonresponse Bias

2. Refusal rate

3. Response Bias

Nonresponse Bias
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Lisa Hardin
2022-07-23

Average
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posader86
2022-07-23

Average
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Noelanijd
2022-07-22

Types Of Research Studies
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Nash Frank
2022-07-21

I started learning series in calculus and I have trouble catching a basic concept. When I try to find if a series converges or diverges I have many ways to go about it. If I see that the series diverges than I stop there. If I see that the series converges than there is a number it's converging to right?

For example: $\sum \frac{2}{{n}^{3}+4}$. I do the limit comparison test with the series $\sum \frac{1}{{n}^{3}}$ and get a finite number 2. I know that $\sum \frac{1}{{n}^{3}}$ converges, so now I know that $\sum \frac{2}{{n}^{3}+4}$ converges also. How do I know to what number it converges to?

Types of Bias
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Brenton Dixon
2022-07-21

$6$ normal dice. The probability to get a $6$ is $1/6$ for each dice.

$3$ biased dice. The probability to get a $6$ is $0.85$.$3$ biased dice. The probability to get a $6$ is $0.05$.You take a die from the box at random and roll it.What is the conditional probability that it is of type $b$, given that it gives a $6$?

Nonresponse Bias
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Livia Cardenas
2022-07-19

1.Respondents have strong opinions about the subject matter.

2.Respondents change their answers to influence the results.

3.Respondents give the answer they think the questioner wants.

4.Respondents do not give their honest opinion.

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Nathalie Fields
2022-07-17

The standart deviation, the arithmetic average and the number of values.

Types Of Research Studies
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Tirimwb
2022-07-14

Suppose a billion monkeys type on word processors at a rate of 10 symbols per second. Assume that the word processors produce 27 symbols, namely, 26 letters of the English alphabet and a space. These monkeys type for 10 billion years. What is the probability that they can type the first sentence of Lincoln’s “Gettysburg Address”?

Four score and seven years ago our fathers brought forth on this continent a new nation conceived in liberty and dedicated to the proposition that all men are created equal.

Hint: Look up Boole’s inequality to provide an upper bound for the probability!

This is a homework question. I just want some pointers how to move forward from what I have done so far. Below I will explain my research so far.

First I calculated the probability of the monkey 1 typing the sentence (this question helped me do that); let's say that probability is p:

$P(\text{Monkey 1 types our sentence})=P({M}_{1})=p$

Now let's say that the monkeys are labeled ${M}_{1}$ to ${M}_{{10}^{9}}$, so given the hint in the question I calculated the upper bound for the probabilities of union of all $P({M}_{i})$ (the probability that i-th monkey types the sentence) using Boole's inequality.

Since $P({M}_{i})=P({M}_{1})=p$,

$P\left(\bigcup _{i}{M}_{i}\right)\le \sum _{i=1}^{{10}^{9}}P({M}_{i})=\sum ^{{10}^{9}}p={10}^{9}\phantom{\rule{thinmathspace}{0ex}}p$

Am I correct till this point? If yes, what can I do more in this question? I tried to study Bonferroni inequality for lower bounds but was unsuccessful to obtain a logical step. If not, how to approach the problem?

Types Of Research Studies
Answered

Elianna Lawrence
2022-07-14

I encountered a set of problems while studying statistics for research which I have combined to get a broader question. I want to know if this is a solvable problem with enough information specifically under what assumptions or approach.

Given that a minesweeper has encountered exactly 5 landmines in a particular 10 mile stretch, what is the probability that he will encounter exactly 6 landmines on the next 10 mile stretch. (Average number of landmines is 0.6 per mile in the 50 mile stretch)

I have figured that the approach involves finding out the Poisson probabilities of the discrete random variable with the combination of Bayes Conditional probability. But am stuck with proceeding on applying the Bayes rule. i.e $Pr(X=6\mid X=5)$.

I know that $Pr(X=5)={e}^{-6}{5}^{6}/5!.$ Here $\lambda =0.6\cdot 10$ and $X=5$) Similarly for $Pr(X=6)$. Is Bayes rule useful here: $P(Y\mid A)=Pr(A\mid Y)Pr(Y)/(Pr(A\mid Y)Pr(Y)+Pr(A\mid N)Pr(N))$?

Would appreciate any hints on proceeding with these types of formulations for broadening my understanding.