10 and np(1-p)>10. In a sample of 100 Americans, 87 were right handed. This means we get started with a set level of confidence and margin of error. This section provides the power calculation formulas for the various test statistics available in this procedure. In order to consider a normal distribution or normal approximation, a standard scale or standard units is necessary. With the classical 30 degrees of freedom the visualization shows that p-value from the normal approximation (0.05) is really close to the p-value from the t-distribution (0.055). The binomial distribution has a mean of $\mu = \text{Np} = 10\cdot 0.5 = 5$ and a variance of $\sigma^2 = \text{Np}(1-\text{p}) = 10 \cdot 0.5\cdot 0.5 = 2.5$; therefore a standard deviation of 1.5811. This is exactly what he did, and the curve he discovered is now called the normal curve. The normal distribution is a good approximation to the binomial when n is sufficiency large and p is not too close to 0 or 1. The hypothesized value of the population proportion is symbolized by $$p_0$$ because this is the value in the null hypothesis ($$H_0$$). Most statistical procedures for testing differences between means assume normal distributions. As probability and statistical theory show us, as the number of samples increase for the given mean and standard deviation, the more closely the sample probability distribution will â¦ Compute a 95% confidence interval to estimate the proportion of all African American adults who have some level of lactose intolerance. The following is an example on how to compute a normal approximation for a binomial distribution. The normal approximation p-value for the three alternative hypotheses uses a â¦ This is a non-directional (i.e., two-tailed) test, so we need to find the area under the z distribution that is more extreme than $$z=-0.980$$. Because there is no estimate of the proportion given, we use $$\tilde{p}=0.50$$ for a conservative estimate. Here, for the sake of ease, we have used an online normal area calculator. giving us an approximation for the variance of our estimator. Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%? The sample observations are not a random sample, so a test about a population proportion using the normal approximating method cannot be used. David Lane, History of Normal Distribution. In this example a success is defined as answering "yes" to the question "do you own a dog?" Normal distribution integral has no analytical solution. Note that p-values are also symbolized by $$p$$. In order to construct a 95% confidence interval with a margin of error of 4%, we should obtain a sample of at least $$n=601$$. Are any of the three requirements violated? k 1.5 Example: Approximate Mean and Variance Suppose X is a random variable with EX = 6= 0. One sample proportion tests and confidence intervals are covered in Section 6.1 of the Lock5 textbook. In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50. Requirements for using normal approximation to binomial. David Lane, History of Normal Distribution. On the following pages you will see how a confidence interval for a population proportion can be constructed by hand using the normal approximation method. $$SE=\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}=\sqrt{\frac{0.640 (1-0.640)}{1168}}=0.014$$, The $$z^*$$ multiplier for a 95% confidence interval is 1.960, The formula for a confidence interval for a proportion is $$\widehat{p}\pm z^* (SE)$$, $$0.640\pm 1.960(0.014)=0.640\pm0.028=[0.612, \;0.668]$$. $$\tilde p$$ is an estimated value of the proportion. It requires knowing the population parameters, not the statistics of a sample drawn from the population of interest. By using regression analysis and after rounding the coefficient to one decimal place, the approximation obtained is () 1 .2 1 .3 5 1 0 .5 Î¦ z = â e â z. Research question: Is this city’s proportion of overweight individuals different from 0.690? The problem is that the binomial distribution is a discrete probablility distribution whereas the normal distribultion is a continuous distribution. $$np_0 = 226(0.50)=113$$ and $$n(1-p_0) = 226(1-0.50)=113$$. The z test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. We can use the normal approximation method. This is a right-tailed test because we want to know if the proportion is greater than 0.80. central limit theorem : The theorem that states: If the sum of independent identically distributed random variables has a finite variance, then it will be (approximately) normally distributed. where $\text{x}$ is the number of heads (60), $\text{N}$ is the number of flips (100), and $\pi$ is the probability of a head (0.5). The binomial distribution can be used to solve problems such as, “If a fair coin is flipped 100 times, what is the probability of getting 60 or more heads?” The probability of exactly $\text{x}$ heads out of $\text{N}$ flips is computed using the formula: $\displaystyle \text{P}\left( \text{x} \right) =\frac { \text{N}! If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. Normalization can also refer to the creation of shifted and scaled versions of statistics, where the intention is that these normalized values allow the comparison of corresponding normalized values for different datasets. np = 583,333.333 >> 10 CHECK! Sum of many independent 0/1 components with probabilities equal p (with n large enough such that npq â¥ 3), then the binomial number of success in n trials can be approximated by the Normal distribution with mean µ = np and standard deviation q np(1âp). One of the conditions for a binomial distribution are not satisfied, so a test about a population proportion using the normal approximating method cannot be used. Find a [latex]\text{Z}$ score for 7.5 using the formula $\text{Z}=\frac { 7.5-5 }{ 1.5811 } =1.5811$. Recall, the z distribution is a normal distribution with a mean of 0 and standard deviation of 1. We are 99% confidence that between 60.4% and 67.6% of all American adults are not financially prepared for retirement. Because the distribution of means is very close to normal, these tests work well even if the distribution itself is only roughly normal. Minitab Express will not check assumptions for you. a. There is no principle that a small number of observations will coincide with the expected value or that a streak of one value will immediately be “balanced” by the others. The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course. When we're constructing confidence intervals $$p$$ is typically unknown, in which case we use $$\widehat{p}$$ as an estimate of $$p$$. The central limit theorem has a number of variants. This means that our sample needs to have at least 10 "successes" and at least 10 "failures" in order to construct a confidence interval using the normal approximation method. You first learned how to construct a frequency table in Lesson 2.1.1.2.1 of these online notes. This is known as the exact method. Because both $$n \widehat p \geq 10$$ and $$n(1- \widehat p) \geq 10$$, the normal approximation method may be used. If the assumptions for the normal approximation method are not met (i.e., if $$np$$ or $$n(1-p)$$ is not at least 10), then the sampling distribution may be approximated using a binomial distribution. According to the Rule of Sample Proportions, if $$np\geq 10$$ and $$n(1-p) \geq 10$$ then the sampling distributing will be approximately normal. We are 95% confident that between 61.2% and 66.8% of all American adults are not financially prepared for retirement. What if we knew that the population proportion was around 0.25? From the Minitab Express output, $$z$$ = -1.86, From the Minitab Express output, $$p$$ = 0.0625, $$p > \alpha$$, fail to reject the null hypothesis. Hamlet Killing Claudius Quotes, Vector Fonts Generator, Bedroom Carpet And Paint Ideas, Andhra And Telangana Cuisine, Do Mountain Goats Attack Humans, Lumix Gf10 Review, " />
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This leads to wider intervals for higher confidence levels. This is known as a normal approximation confidence interval. A problem arises when there are a limited number of samples, or draws in the case of data “drawn from a box.” A probability histogram of such a set may not resemble the normal curve, and therefore the normal curve will not accurately represent the expected values of the random variables. The 99% confidence interval will be wider than the 95% confidence interval. However, knowing the true standard deviation of a population is often unrealistic except in cases such as standardized testing, where the entire population is measured. A hypothesis test for a proportion is used when you are comparing one group to a known or hypothesized population proportion value. This can be done using raw data or summarized data. $\begingroup$ One advantage of using the normal is it often gives enough information to quickly tell whether it's even worth calculating the answer more precisely. This is a left-tailed test because we want to know if the proportion is less than 0.80. The process of using this curve to estimate the shape of the binomial distribution is known as normal approximation. Both $$np_0 \geq 10$$ and $$n(1-p_0) \geq 10$$ so we can use the normal approximation method. The runs test is a useful tool to determine if a sequence is likely to be random or not. All of the conditions for testing a claim about a population proportion using the normal approximation method are satisfied, so the method can be used. The normal approximation can be used in counting problems, where the central limit theorem includes a discrete-to-continuum approximation and where infinitely divisible and decomposable distributions are involved. Recall that if $$np \geq 10$$ and $$n(1-p) \geq 10$$ then the sampling distribution can be approximated by a normal distribution. TODO: binom_test intervals raise an exception in small samples if one. This method of constructing a sampling distribution is known as the normal approximation method. You can change this value by clicking on the distributions. A key point is that calculating $\text{z}$ requires the population mean and the population standard deviation, not the sample mean or sample deviation. $$H_{a}\colon p>0.80$$, $$z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}$$, $$\widehat{p}=\dfrac{87}{100}=0.87$$, $$p_{0}=0.80$$, $$n=100$$, $$z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.87-0.80}{\sqrt{\frac{0.80 (1-0.80)}{100}}}=1.75$$. Since p is close to ½ (it equals ½! Method âbinom_testâ directly inverts the binomial test in scipy.stats. It is important to remember that the LLN only applies (as the name indicates) when a large number of observations are considered. What if we have summarized data and not data in a Minitab Express worksheet? In more complicated cases, normalization may refer to more sophisticated adjustments where the intention is to bring the entire probability distributions of adjusted values into alignment. Of the 522 students in the sample, 273 said that they did have a dog. Asked Jul 19, 2020. The scope of the normal approximation is dependent upon our sample size, becoming more accurate as the sample size grows. To calculate this area, first we compute the area below 8.5 and then subtract the area below 7.5. In a representative sample of 1168 American adults, 747 said they were not financially prepared for retirement. A function of the form Î¦(z )= 1 â 0 .5 e â Az b can be used as an approximation to the standard normal cumulative function. According to the law of large numbers, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. Both $$n p_0$$ and $$n (1-p_0)$$ are at least 10, this assumption has been met. O B. This means we can use the normal approximation method to construct this confidence interval. When Is the Approximation Appropriate? $$p_{0}$$ = hypothesize population proportion This must be done manually. Find a $\text{Z}$ score for 8.5 using the formula $\text{Z}=\frac { 8.5-5 }{ 1.5811 } =2.21$. Therefore, de Moivre reasoned that if he could find a mathematical expression for this curve, he would be able to solve problems such as finding the probability of 60 or more heads out of 100 coin flips much more easily. When discussion proportions, we sometimes refer to this as the Rule of Sample Proportions. The results are shown in the following figures: Normal Area 2: This graph shows the area below 7.5. normal approximation: The process of using the normal curve to estimate the shape of the distribution of a data set. The standard score is the number of standard deviations an observation or datum is above the mean. In Minitab Express, the exact method is the default method. The value of the multiplier increases as the confidence level increases. In order to use the normal approximation method, the assumption is that both $$n p_0 \geq 10$$ and $$n (1-p_0) \geq 10$$. There were 24 females. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. The tool of normal approximation allows us to approximate the probabilities of random variables for which we don’t know all of the values, or for a very large range of potential values that would be very difficult and time consuming to calculate. We want to construct a 95% confidence interval for $$p$$ with a margin of error equal to 4%. Using Minitab Express, we find the probability $$P(z\geq1.75)=0.0400592$$ which may be rounded to $$p\; value=0.0401$$. We want to construct a 95% confidence interval for $$p$$ with a margin of error equal to 4%. Laplace showed that even if a distribution is not normally distributed, the means of repeated samples from the distribution would be very nearly normal, and that the the larger the sample size, the closer the distribution would be to a normal distribution. A normal distribution has some interesting properties: it has a bell shape, the mean and median are equal, and 68% of the data falls within 1 standard deviation. X ~ N(20 × ½, 20 × ½ × ½) so X ~ N(10, 5) . In probability theory and statistics, the chi-square distribution (also chi-squared or Ï 2-distribution) with k degrees of freedom is the distribution of a sum of the squares of k independent standard normal random variables. The importance of the normal curve stems primarily from the fact that the distribution of many natural phenomena are at least approximately normally distributed. If you have data in a Minitab Express worksheet, then you have what we call "raw data." The central limit theorem (CLT) states that, given certain conditions, the mean of a sufficiently large number of independent random variables, each with a well-defined mean and well-defined variance, will be approximately normally distributed. Minitab evaluates the likelihood ratio for all possible values of X = (0, 1,â¦, n) and sums the probabilities for all values for which the LR (y) â¥ LR (x).. p-value = Î£ P{X = y | â¦ This same distribution had been discovered by Laplace in 1778—when he derived the extremely important central limit theorem. In Spring 2016, a sample of 522 World Campus students were surveyed and asked if they own a dog. Some types of normalization involve only a rescaling, to arrive at values relative to some size variable. $$99\%\;C.I. The question then is, “What is the probability of getting a value exactly 1.8973 standard deviations above the mean?” You may be surprised to learn that the answer is 0 (the probability of any one specific point is 0). Abraham de Moivre, an 18th century statistician and consultant to gamblers, was often called upon to make these lengthy computations. This is in contrast to "summarized data" which you'll see on the next page. Nevertheless, there are several methods which provide an approximation of the integral by numerical methods: Taylor series, asymptotic series, continual fractions, and some other more. Normal Area 1: This graph shows the area below 8.5. Where \(p_0$$ is the hypothesized population proportion that you are comparing your sample to. The binomial distribution has a mean of $\mu = \text{Np} = 10\cdot 0.5 = 5$ and a variance of $\sigma^2 = \text{Np}(1-\text{p}) = 10 \cdot 0.5\cdot 0.5 = 2.5$. Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. That sampling distribution will have a mean of $$p_0$$ and a standard deviation (i.e., standard error) of $$\sqrt{\frac{p_0 (1-p_0)}{n}}$$, Recall that the standard normal distribution is also known as the z distribution. Note that the default method for constructing the sampling distribution in Minitab Express is to use the exact method. There is evidence that the proportion of women in the population who think they are overweight is less than 40%. Normal Distribution and Scales: Compares the various grading methods in a normal distribution. 1. where p = proportion of interest 2. n = sample size 3. $$H_{0}\colon p=0.80$$ which has discrete steps. is approximately distributed as a normal distribution with a mean of 0 and a standard deviation of 1, N(0,1). Question 1. 1. Testing the Normal Approximation and Minimal Sample Size Requirements of Weighted Kappa When the Number of Categories is Large Domenic V. Cicchetti Applied Psychological Measurement 1981 5 : â¦ Normal Approximation: The normal approximation to the binomial distribution for 12 coin flips. A key point is that calculating $\text{z}$ requires the population mean and the population standard deviation, not the sample mean or sample deviation. This course does not cover the exact method in detail, but you will see how these tests may be performed using Minitab Express. We can use these pieces to determine a minimum sample size needed to produce these results by using algebra to solve for $$n$$: $$M$$ is the margin of error Can a test about a population proportion using the normal approximation method be used? Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. From the plot below, we see that the $$z^*$$ multiplier for a 99% confidence interval is 2.576. Let's construct a 95% confidence interval to estimate the proportion of all American adults who are not financially prepared for retirement. Normal Approximation Method Power may be calculated for one-sample proportions tests using the normal approximation to the binomial distribution. Translate the problem into a probability statement about X. In variants, convergence of the mean to the normal distribution also occurs for non-identical distributions, given that they comply with certain conditions. Here is another example: To create a frequency table of dog ownership in Minitab Express: This should result in the following frequency table: Select your operating system below to see a step-by-step guide for this example. did not choose Normal Approximation as the method)? A failure is defined as answering "no." If one only has a sample set, then the analogous computation with sample mean and sample standard deviation yields the Student’s $\text{t}$-statistic. September 17, 2013. According to the Center for Disease Control (CDC), the percent of adults 20 years of age and over in the United States who are overweight is 69.0% (see http://www.cdc.gov/nchs/fastats/obesity-overweight.htm). To perform a one sample proportion z test with summarized data in Minitab Express: $$p \leq \alpha$$, reject the null hypothesis. September 17, 2013. $$H_{0}\colon p=0.80$$ (adsbygoogle = window.adsbygoogle || []).push({}); The process of using the normal curve to estimate the shape of the binomial distribution is known as normal approximation. The $$z^*$$ multiplier for a 95% confidence interval is 1.960. As shown on the probability distribution plot below, the multiplier associated with a 95% confidence interval is 1.960, often rounded to 2 (recall the Empirical Rule and 95% Rule). A. Check rules of thumb using n = 3,500,000 and p = 1/6. Now, we have an estimate to include in the formula: $$n=\left ( \frac{1.960}{0.04} \right )^2 (0.25)(1-0.25)=450.188$$. The standard score is a dimensionless quantity obtained by subtracting the population mean from an individual raw score and then dividing the difference by the population standard deviation. :\;0.640\pm 2.576 (0.014)=0.0640\pm 0.036=[0.604, \; 0.676]\). When conducting a hypothesis test, we check this assumption using the hypothesized proportion (i.e., the proportion in the null hypothesis). This is exactly what he did, and the curve he discovered is now called the normal curve. Verify whether n is large enough to use the normal approximation by checking the two appropriate conditions.. For the above coin-flipping question, the conditions are met because n â p = 100 â 0.50 = 50, and n â (1 â p) = 100 â (1 â 0.50) = 50, both of which are at least 10.So go ahead with the normal approximation. In cases where it is impossible to measure every member of a population, a random sample may be used. The equation for the Normal Approximation for the Binomial CI is shown below. Clearly, the normal approximation to the binomial is a much better method. }{ \text{x}!\left( \text{N}-\text{x} \right) ! } Central Limit Theorem: A distribution being “smoothed out” by summation, showing original density of distribution and three subsequent summations. Normal Approximation 3 of6 0 5 10 15 20 25 30 0.00 0.05 0.10 0.15 Normal Approx to Binom: n=20, p=0.5 x binomial dist P(x) normal approx f(x) Thus, if np 5 and nq 5 we can use the normal distribution to approxi-mately describe a binomial random variable. David Lane, Normal Approximation to the Binomial. In other words, the scope of the normal approximation is dependent upon our sample size, becoming more accurate as the sample size grows. Check assumptions and write hypotheses, 3. Thus, this is known as a "single sample proportion z test" or "one sample proportion z test.". The next two pages will show you how to use Minitab Express to conduct this analysis using either raw data or summarized data. Research Question: Are more than 80% of American's right handed? The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. $$n$$ = sample size. Note: Because the normal approximation is not accurate for small values of n, a good rule of thumb is to use the normal approximation only if np>10 and np(1-p)>10. In a sample of 100 Americans, 87 were right handed. This means we get started with a set level of confidence and margin of error. This section provides the power calculation formulas for the various test statistics available in this procedure. In order to consider a normal distribution or normal approximation, a standard scale or standard units is necessary. With the classical 30 degrees of freedom the visualization shows that p-value from the normal approximation (0.05) is really close to the p-value from the t-distribution (0.055). The binomial distribution has a mean of $\mu = \text{Np} = 10\cdot 0.5 = 5$ and a variance of $\sigma^2 = \text{Np}(1-\text{p}) = 10 \cdot 0.5\cdot 0.5 = 2.5$; therefore a standard deviation of 1.5811. This is exactly what he did, and the curve he discovered is now called the normal curve. The normal distribution is a good approximation to the binomial when n is sufficiency large and p is not too close to 0 or 1. The hypothesized value of the population proportion is symbolized by $$p_0$$ because this is the value in the null hypothesis ($$H_0$$). Most statistical procedures for testing differences between means assume normal distributions. As probability and statistical theory show us, as the number of samples increase for the given mean and standard deviation, the more closely the sample probability distribution will â¦ Compute a 95% confidence interval to estimate the proportion of all African American adults who have some level of lactose intolerance. The following is an example on how to compute a normal approximation for a binomial distribution. The normal approximation p-value for the three alternative hypotheses uses a â¦ This is a non-directional (i.e., two-tailed) test, so we need to find the area under the z distribution that is more extreme than $$z=-0.980$$. Because there is no estimate of the proportion given, we use $$\tilde{p}=0.50$$ for a conservative estimate. Here, for the sake of ease, we have used an online normal area calculator. giving us an approximation for the variance of our estimator. Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%? The sample observations are not a random sample, so a test about a population proportion using the normal approximating method cannot be used. David Lane, History of Normal Distribution. In this example a success is defined as answering "yes" to the question "do you own a dog?" Normal distribution integral has no analytical solution. Note that p-values are also symbolized by $$p$$. In order to construct a 95% confidence interval with a margin of error of 4%, we should obtain a sample of at least $$n=601$$. Are any of the three requirements violated? k 1.5 Example: Approximate Mean and Variance Suppose X is a random variable with EX = 6= 0. One sample proportion tests and confidence intervals are covered in Section 6.1 of the Lock5 textbook. In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50. Requirements for using normal approximation to binomial. David Lane, History of Normal Distribution. On the following pages you will see how a confidence interval for a population proportion can be constructed by hand using the normal approximation method. $$SE=\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}=\sqrt{\frac{0.640 (1-0.640)}{1168}}=0.014$$, The $$z^*$$ multiplier for a 95% confidence interval is 1.960, The formula for a confidence interval for a proportion is $$\widehat{p}\pm z^* (SE)$$, $$0.640\pm 1.960(0.014)=0.640\pm0.028=[0.612, \;0.668]$$. $$\tilde p$$ is an estimated value of the proportion. It requires knowing the population parameters, not the statistics of a sample drawn from the population of interest. By using regression analysis and after rounding the coefficient to one decimal place, the approximation obtained is () 1 .2 1 .3 5 1 0 .5 Î¦ z = â e â z. Research question: Is this city’s proportion of overweight individuals different from 0.690? The problem is that the binomial distribution is a discrete probablility distribution whereas the normal distribultion is a continuous distribution. $$np_0 = 226(0.50)=113$$ and $$n(1-p_0) = 226(1-0.50)=113$$. The z test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. We can use the normal approximation method. This is a right-tailed test because we want to know if the proportion is greater than 0.80. central limit theorem : The theorem that states: If the sum of independent identically distributed random variables has a finite variance, then it will be (approximately) normally distributed. where $\text{x}$ is the number of heads (60), $\text{N}$ is the number of flips (100), and $\pi$ is the probability of a head (0.5). The binomial distribution can be used to solve problems such as, “If a fair coin is flipped 100 times, what is the probability of getting 60 or more heads?” The probability of exactly $\text{x}$ heads out of $\text{N}$ flips is computed using the formula: $\displaystyle \text{P}\left( \text{x} \right) =\frac { \text{N}! If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. Normalization can also refer to the creation of shifted and scaled versions of statistics, where the intention is that these normalized values allow the comparison of corresponding normalized values for different datasets. np = 583,333.333 >> 10 CHECK! Sum of many independent 0/1 components with probabilities equal p (with n large enough such that npq â¥ 3), then the binomial number of success in n trials can be approximated by the Normal distribution with mean µ = np and standard deviation q np(1âp). One of the conditions for a binomial distribution are not satisfied, so a test about a population proportion using the normal approximating method cannot be used. Find a [latex]\text{Z}$ score for 7.5 using the formula $\text{Z}=\frac { 7.5-5 }{ 1.5811 } =1.5811$. Recall, the z distribution is a normal distribution with a mean of 0 and standard deviation of 1. We are 99% confidence that between 60.4% and 67.6% of all American adults are not financially prepared for retirement. Because the distribution of means is very close to normal, these tests work well even if the distribution itself is only roughly normal. Minitab Express will not check assumptions for you. a. There is no principle that a small number of observations will coincide with the expected value or that a streak of one value will immediately be “balanced” by the others. The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course. When we're constructing confidence intervals $$p$$ is typically unknown, in which case we use $$\widehat{p}$$ as an estimate of $$p$$. The central limit theorem has a number of variants. This means that our sample needs to have at least 10 "successes" and at least 10 "failures" in order to construct a confidence interval using the normal approximation method. You first learned how to construct a frequency table in Lesson 2.1.1.2.1 of these online notes. This is known as the exact method. Because both $$n \widehat p \geq 10$$ and $$n(1- \widehat p) \geq 10$$, the normal approximation method may be used. If the assumptions for the normal approximation method are not met (i.e., if $$np$$ or $$n(1-p)$$ is not at least 10), then the sampling distribution may be approximated using a binomial distribution. According to the Rule of Sample Proportions, if $$np\geq 10$$ and $$n(1-p) \geq 10$$ then the sampling distributing will be approximately normal. We are 95% confident that between 61.2% and 66.8% of all American adults are not financially prepared for retirement. What if we knew that the population proportion was around 0.25? From the Minitab Express output, $$z$$ = -1.86, From the Minitab Express output, $$p$$ = 0.0625, $$p > \alpha$$, fail to reject the null hypothesis. 