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2022-07-22
Answered

How To Calculate Standard Deviation??

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Cheyanne Charles

Answered 2022-07-23
Author has **13** answers

First, you need to determine the mean. The mean of a list of numbers is the sum of those numbers divided by the quantity of items in the list (read: add all the numbers up and divide by how many there are).

Then, subtract the mean from every number to get the list of deviations. Create a list of these numbers. It's OK to get negative numbers here. Next, square the resulting list of numbers (read: multiply them with themselves).

Add up all of the resulting squares to get their total sum. Divide your result by one less than the number of items in the list.

To get the standard deviation, just take the square root of the resulting number

I know this sounds confusing, but just check out this example:

your list of numbers: 1, 3, 4, 6, 9, 19

mean:$\frac{(1+3+4+6+9+19)}{6}=\frac{42}{6}=7$

list of deviations: -6, -4, -3, -1, 2, 12

squares of deviations: 36, 16, 9, 1, 4, 144

sum of deviations: 36+16+9+1+4+144 = 210

divided by one less than the number of items in the list: $\frac{210}{5}=42$

square root of this number: square $\sqrt{42}$ = about 6.48

Then, subtract the mean from every number to get the list of deviations. Create a list of these numbers. It's OK to get negative numbers here. Next, square the resulting list of numbers (read: multiply them with themselves).

Add up all of the resulting squares to get their total sum. Divide your result by one less than the number of items in the list.

To get the standard deviation, just take the square root of the resulting number

I know this sounds confusing, but just check out this example:

your list of numbers: 1, 3, 4, 6, 9, 19

mean:$\frac{(1+3+4+6+9+19)}{6}=\frac{42}{6}=7$

list of deviations: -6, -4, -3, -1, 2, 12

squares of deviations: 36, 16, 9, 1, 4, 144

sum of deviations: 36+16+9+1+4+144 = 210

divided by one less than the number of items in the list: $\frac{210}{5}=42$

square root of this number: square $\sqrt{42}$ = about 6.48

asked 2021-10-14

How can the normal distribution be used to test hypotheses about populations? Also, how might you surmise that you can use this logic to test a hypothesis such as caffeine improves memory in the caffeine study.

asked 2021-06-07

Smoking and drinking coffee have a tendency to stain teeth. In an effort to determine the ability of chewing gum to remove stains on teeth, researchers conducted an experiment in which 64 bovine incisors (teeth) were stained with natural pigments such as coffee for 10 days. Each tooth was randomly assigned to one of four treatments: gum A, gum B, gum C, or saliva. Ea ch tooth group was placed into a device that simulated a human chewing gum. The temperature of the device was maintained at body temperature and the tooth was in the device for 20 minutes. The process was repeated six times (for a total of 120 minutes of chewing). The researcher conducting the experiment did not know which treatment was being applied to each experimental unit. Upon removing a tooth from the chewing apparatus, the color change was measured using a spectrophotometer. The percent age of stain removed by each treatment after 120 minutes is as follows: Gum A, 47.6%; Gum B, 45.2%, Gum C, 21.4%, Saliva, 2.1%. The researchers concluded that gums A and B removed significantly more stain than gum C or saliva. In addition, gum C removed significantly more stain than saliva. (a) Identify the research objective. (b) Is this an observational study or designed experiment? Why? (c) If observational, what type of observational study is this? If an experiment, what type of experimental design is this? (d) What is the response variable? (e) What is the explanatory variable? ls it qualitative or quantitative? (I) Identify the experimental units. (g) State a factor that could affect the value of the response variable that is fixed at a set level. (h) What is the conclusion of the study?

asked 2022-05-27

I have "solved" a question in mathematical statistics, but unfortunately received an incorrect answer and I would appreciate any help in understanding why my method did not work in this case. The question is as follows:

In city A, 500 randomly selected people were tested for antibodies to covid-19, which resulted in 110 giving positive results. In city B, a random sample of 500 people was also drawn, and here 91 had positive results. Can it be said that it is statistically certain that city A has a larger population share with antibodies than city B? (Justify your answer with an appropriate test. Use 5% significance level.)

I started by producing ${P}^{\ast}(A)=\frac{110}{500}=0.22$ and respectively ${P}^{\ast}(B)=\frac{91}{500}=0.182$.

Then I chose to continued with a one-sided hypothesis testing where I assumed that the data was binomially distributed according to $Bin(n=500,p)$, with${H}_{0}:P=0.182\text{and}{H}_{1}:\mathrm{0.182.}$

And finally brought out the one-sided interval according to:${I}_{p}={\lambda}_{\alpha}\sqrt{\frac{{P}^{\ast}(A)(1-{P}^{\ast}(A))}{n}}\Rightarrow [0.189,\mathrm{\infty}].$.And concluded that ${H}_{0}$ should be rejected since 0.182 is not included in the interval. But according to the results, ${H}_{0}$ should not be rejected and they have also used a different method than I did. Why is this method not applicable in my case?

Note: At first I thought that errors were because I only used α and not $\frac{\alpha}{2}$ in the calculation, however, even if I were to replace this, 0.182 is not in the interval.

In city A, 500 randomly selected people were tested for antibodies to covid-19, which resulted in 110 giving positive results. In city B, a random sample of 500 people was also drawn, and here 91 had positive results. Can it be said that it is statistically certain that city A has a larger population share with antibodies than city B? (Justify your answer with an appropriate test. Use 5% significance level.)

I started by producing ${P}^{\ast}(A)=\frac{110}{500}=0.22$ and respectively ${P}^{\ast}(B)=\frac{91}{500}=0.182$.

Then I chose to continued with a one-sided hypothesis testing where I assumed that the data was binomially distributed according to $Bin(n=500,p)$, with${H}_{0}:P=0.182\text{and}{H}_{1}:\mathrm{0.182.}$

And finally brought out the one-sided interval according to:${I}_{p}={\lambda}_{\alpha}\sqrt{\frac{{P}^{\ast}(A)(1-{P}^{\ast}(A))}{n}}\Rightarrow [0.189,\mathrm{\infty}].$.And concluded that ${H}_{0}$ should be rejected since 0.182 is not included in the interval. But according to the results, ${H}_{0}$ should not be rejected and they have also used a different method than I did. Why is this method not applicable in my case?

Note: At first I thought that errors were because I only used α and not $\frac{\alpha}{2}$ in the calculation, however, even if I were to replace this, 0.182 is not in the interval.

asked 2022-03-23

Given $X}_{1},\dots ,{X}_{100$ and $Y}_{1},\dots ,{Y}_{50$ which are independent random samples from identical distribution $N(\mu ,1)$ . Each of two statisticians is trying to build confidence interval for $\mu$ on a confidence level of 0.8 $1-\alpha =0.8$ . You have to find probability that those intervals are disjoint. I know i have to find $P(|\stackrel{\u2015}{X}-\stackrel{\u2015}{Y}|>\frac{1\cdot 1.28}{\sqrt{50}}+\frac{1\cdot 1.28}{\sqrt{100}})$ which is $1-P(-0.309\le \stackrel{\u2015}{X}-\stackrel{\u2015}{Y}\le 0.309)$

Question is what should i do next how to calculate this probability?

Question is what should i do next how to calculate this probability?

asked 2022-07-07

The data generating process is

$f(x\mid \theta )=\frac{1}{\theta},\phantom{\rule{1em}{0ex}}0\le x\le \theta $

Let ${H}_{1}:\theta =5$ and ${H}_{2}:\theta \ne 5$. Prior probabilities of $\frac{1}{2}$ are assigned to the two hypotheses, and a sample of size $5$ yields a maximum value of $4$. Find the $p$-value here.

According to the definition of $p$-value,

$p=\mathbb{P}(LR\le L{R}^{\ast}\mid {H}_{1}),$

and $L{R}^{\ast}={\left(\frac{4}{5}\right)}^{5}$. I am not sure what to do next. Furthermore, intuitively, if the maxima is equal to $5$ exactly, shall we get an extremely big $p$ or a small one?

$f(x\mid \theta )=\frac{1}{\theta},\phantom{\rule{1em}{0ex}}0\le x\le \theta $

Let ${H}_{1}:\theta =5$ and ${H}_{2}:\theta \ne 5$. Prior probabilities of $\frac{1}{2}$ are assigned to the two hypotheses, and a sample of size $5$ yields a maximum value of $4$. Find the $p$-value here.

According to the definition of $p$-value,

$p=\mathbb{P}(LR\le L{R}^{\ast}\mid {H}_{1}),$

and $L{R}^{\ast}={\left(\frac{4}{5}\right)}^{5}$. I am not sure what to do next. Furthermore, intuitively, if the maxima is equal to $5$ exactly, shall we get an extremely big $p$ or a small one?

asked 2021-10-27

Trucks are used in many configurations. Three common configurations are tractors with 1, 2, and 3 trailers. Accident statistics for these configurations in 2002 are listed below.

Fatal U.S. Truck Accidents in 2002 by Configuration

Assume these statistics represent simple random samples for each configuration. Which of the following would be the most, appropriate test to use to determine whether the populations for each configuration have the same rate of fatal accidents?

A) One-sample proportion z-test.

B) Two-sample proportion z-test

C) Chi-square test for independence

D) Chi-square goodness-of-fit test

E) Chi-square test for homogeneity of proportions

asked 2022-06-19

What is the geometric mean of 3,6 and 12?