# For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 Leftarrow p Leftarrow 0.95), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation sigma = sqrt{np(1 - p)}. Explain this approximation making use the Central Limit Theorem. Question
Transformation properties For large value of n, and moderate values of the probabiliy of success p (roughly, $$0.05\ \Leftarrow\ p\ \Leftarrow\ 0.95$$), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation
$$\sigma = \sqrt{np(1\ -\ p)}$$. Explain this approximation making use the Central Limit Theorem. 2021-03-19
Step 1
Alternatively prove that:
Theorem:
If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the
variablez $$=\frac{x\ -\ \mu}{\sigma}\ -\ N(0,\ 1)$$
where $$\mu=np\ \sigma^{2}=np(1\ -\ p)$$
Proof:
It can be prove using Moment generating function for binomial distribution. It's given as,
$$M_{x}(\theta)=(q\ +\ pe^{\theta})^{n}$$
where $$q = 1\ -\ p.$$
Step 2
By the linear transformation properties of the moment generating function.
$$M_{z}(\theta)=e\ -\ \frac{\mu\ \theta}{\sigma}M_{x}\left(\frac{\theta}{\sigma}\right)=e\ -\ \frac{\mu\ \theta}{\sigma}\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)^{n}$$
Taking the natural log of both sides, and then expanding the power series of $$e^{\theta/\sigma}$$
Then, $$\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)$$
Since $$p\ +\ q = 1.$$
$$=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(1\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)$$
If n is made sufficiently large $$\sigma=\sqrt{npq}$$ can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.
Let $$t=p\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}$$
Thus for sufficiently large $$n,\ | t |\ <\ 1.$$</span>
The ln term in the previous expression is $$\ln (1\ +\ t)\ where\ | t |\ <\ 1,$$</span> and so we may expand this term as follows:
$$\ln(1\ +\ t)=\sum_{k=1}^{\infty}(-1)^{k\ -\ 1}\frac{t^{k}}{k}$$
Step 3
This means that
$$\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln(1\ +\ t)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \sum_{k=1}^{\infty}(-1)^{k_{1}}$$
$$\frac{t_{k}}{k}=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\ \sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ -\ p^{2}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{2}\right)+\ p^{3}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right)^{3}\ -\ \cdots\right]=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\left[\frac{\theta}{\sigma}\ +\ \frac{1}{2}\left(\frac{\theta}{\sigma}\right)^{2}\ +\ \sum_{m=3}^{\infty}\ \frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]\ -\ p^{2}\left[\left(\frac{\theta}{\sigma}\right)^{2}\ +\ 2\left(\frac{\theta}{\sigma}\right)\sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ +\ \sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]^{2}\right]$$
+ an infinite series of terms involving $$\left(\frac{\theta}{\sigma}\right)^{m}\ with\ m\ \geq\ 3]$$
By collecting terms in powers of $$\frac{\theta}{\sigma},$$ we see that
$$\ln\ M_{z}(\theta)=(-\mu\ +\ np)\frac{\theta}{\sigma}\ +\ n(p\ -\ p^{2})\left(\frac{\theta}{\sigma}\right)^{2}\ +\ n\ \sum_{m=3}^{\infty}c_{m}\left(\frac{\theta}{\sigma}\right)^{m}$$
$$=\frac{np\ -\ \mu}{\sigma}\theta\ +\ \frac{np(1\ -\ p)}{\sigma^{2}}\theta^{2}\ +\ n\sum_{m=3}^{\infty}\frac{c_{m}}{\sigma_{m}}\theta^{m}$$
Here, the $$c_{k}$$ terms don’t involve n, sigma or theta.
Since \mu=np\ and\ \sigma^{2}=mp(1\ -\ p)\)
the coefficient of the theta term is 0 and the coefficient of the
$$\theta^{2}$$ term is 1. Thus
$$\ln\ M_{z}(\theta)=\ \frac{\theta^{2}}{2}\ +\ \sum_{m=3}^{\infty}\ \frac{nc_{m}}{\sigma^{m}}\theta^{m}$$
Since the coefficient of each term in the sum has form
$$\frac{nc_{m}}{\sigma^{m}}=\frac{nc_{m}}{(npq)^{m/2}}= \frac{c_{m}}{(pq)^{m/2}}\ \cdot\ \frac{1}{n^{m/2\ -\ 1}}\ \rightarrow\ 0asn\ \rightarrow\ \infty$$
$$\lim_{n\ \rightarrow\ \infty}\ \ln\ M_{z}(\theta)=\frac{\theta^{2}}{2}$$
$$\lim_{n\ \rightarrow\ \infty}\ M_{z}(\theta)=e^{\theta2/2}$$
But note that by Property 3 of Normal Distribution the moment generating function for a random variable z with distribution N (0, 1) is
$$M_{z}(\theta)=e^{\theta^{2}/2}$$
Hence , Proved .

### Relevant Questions Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. 1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance. factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution. Which of the following are correct general statements about the Central Limit Theorem?
(Select all that apply. To be marked correct: All of the correct selections must be made, with no incorrect selections.)
Question 3 options:
Its name is often abbreviated by the three capital letters CLT.
The accuracy of the approximation it provides, improves as the sample size n increases.
The word Central within its name, is meant to signify its role of central importance in the mathematics of probability and statistics.
It is a special example of the particular type of theorems in mathematics, which are called Limit Theorems.
It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$.
It specifies the specific shape of the curve which approximates certain sampling distributions.
It specifies the specific mean of the curve which approximates certain sampling distributions. A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value Which of the following are correct general statements about the central limit theorem? Select all that apply
1. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$
2. It specifies the specific mean of the curve which approximates certain sampling distributions.
3. It is a special example of the particular type of theorems in mathematics, which are called Limit theorems.
4. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
5. It’s name is often abbreviated by the three capital letters CLT.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. The word Central within its name, is mean to signify its role of central importance in the mathematics of probability and statistics.
8. It specifies the specific shape of the curve which approximates certain sampling distributions. Which of the following are correct general statements about the Central Limit Theorem? Select all that apply.
1. It specifies the specific shape of the curve which approximates certain sampling distributions.
2. It’s name is often abbreviated by the three capital letters CLT
3. The word Central within its name, is meant to signify its role of central importance in the mathematics of probability and statistics.
4. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to 50\%.
5. It specifies the specific mean of the curve which approximates certain sampling distributions.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
8. It is a special example of the particular type of theorems in mathematics, which are called limit theorems. Which of the following are correct general statements about the central limit theorem? Select all that apply
1. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$
2. It specifies the specific mean of the curve which approximates certain sampling distributions.
3. It is a special example of the particular type of theorems in mathematics, which are called Limit theorems.
4. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
5. It’s name is often abbreviated by the three capital letters CLT.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. The word Central within its name, is mean to signify its role of central importance in the mathematics of probability and statistics.
8. It specifies the specific shape of the curve which approximates certain sampling distributions. A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings. This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function $$p(x)=\frac{1}{\sqrt{2 \pi}^{\sigma}}e^{-(x-\mu)^{2}}/(2 \sigma^{2})$$ where $$\pi = 3.14159265 . . .$$ and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph$$(\text{for}\ \sigma=1\ \text{and}\ \mu=2)$$is shown in the figure. With $$\sigma = 5 \text{and} \mu = 0$$, approximate $$\int_0^{+\infty}\ p(x)\ dx.$$