Sampling Distribution Mean and Standard Error Questions

Question Description

Describe the shape of the sampling distribution of the sample mean and find its mean and standard error 

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ampling Distributions and Large Sample Estimation Department of Statistics MiamiOH.edu/cas @CASMiamiOH 2 What we will Cover โ€ข Sampling Distributions โ€ข The Central Limit Theorem โ€ข Sample Statistic and Their Distributions โ€ข Sample Mean, ๐‘ฅาง โ€ข Sample Proportion, ๐‘ฦธ โ€ข Large Sample Estimation MiamiOH.edu/cas @CASMiamiOH 3 Sampling Distributions โ€ข Numerical descriptive measures calculated from the sample are called statistics โ€ข Example: Sample Mean (๐‘ฅ)าง and Sample Proportion (๐‘)ฦธ โ€ข Statistics are random variables because they vary from sample to sample. โ€ข The probability distributions for statistics are called sampling distributions โ€ข In repeated sampling, they tell us what values of the statistics can occur and how often each value occurs. MiamiOH.edu/cas @CASMiamiOH 4 Sampling Distribution Continued โ€ข Sampling distributions for statistics can be โ€ข Approximated with simulation techniques โ€ข Derived using mathematical theorems

โ€ข The Central Limit Theorem (CLT) is one such theorem. Central Limit Theorem (CLT) : If random samples of ๐‘› observations are drawn from a population with any underlying distribution with a finite ๐œ‡ and standard deviation ๐œŽ. Then 1 when ๐‘› is large, the sampling distribution of the sample mean (เดคx = ฯƒni=1 xi ) is n approximately normally distributed with mean ๐œ‡ and standard deviation ๐œŽ . ๐‘› Notice the approximation becomes more accurate as ๐‘› becomes large. MiamiOH.edu/cas @CASMiamiOH 5 Importance of CLT โ€ข The Central Limit Theorem also implies that the sum of ๐‘› measurements is approximately normal with mean ๐‘›๐œ‡ and standard deviation ๐œŽ ๐‘› ๐ธ ๐‘›๐œ‡ = ๐‘›๐œ‡ and Var[nเดคx] = ๐‘›2 ๐œŽ2 = ๐‘›๐œŽ 2 โ‡’ Std. Dev = ๐‘›๐œŽ 2 = ๐‘›๐œŽ ๐‘› โ€ข Many statistics that are used for statistical inference are sums or average of sample measurements.

โ€ข When ๐‘› is large, these statistics will have approximately normal distribution. โ€ข This will allow us to describe their behavior and evaluate the reliability of our inferences. MiamiOH.edu/cas @CASMiamiOH 6 Relationship of CLT and the Sample Size โ€ข If the sample is normal, then the sampling distribution of ๐‘ฅาง will also be normal, no matter what the sample size is. โ€ข When the sample population is approximately symmetric, the distribution becomes approximately normal even for relatively small values ๐‘›. โ€ข When the sample population is skewed, the sample size must be at least 30 before the sampling distribution ๐‘ฅาง becomes approximately normal. MiamiOH.edu/cas @CASMiamiOH 7 Sampling Distribution of the Sample Mean โ€ข A random sample of size ๐‘› is selected from a population with mean ๐œ‡ and standard deviation ๐œŽ. ๐œŽ ๐‘›

โ€ข The sampling distribution of the sample mean ๐‘ฅาง will have mean ๐œ‡ and โ€ข If the original population is normal, the sampling distribution will be normal for any sample size. โ€ข If the original population is non-normal, the sampling distribution will be normal when ๐‘› is large The standard deviation of ๐‘ฅาง is sometimes called the Standard Error (SE). MiamiOH.edu/cas @CASMiamiOH 8 Probabilities for the Sample Mean โ€ข If the sampling distribution of ๐‘ฅาง is normal or approximately normal, standardize or rescale the interval of interest in terms of ๐‘ง= โ€ข ๐‘ฅาง โˆ’ ๐œ‡ ๐œŽฮค ๐‘› Find the appropriate area using a Z-Score table Example: A random sample of size ๐‘› = 16 from a normal distribution with ๐œ‡ = 10 and ๐œŽ = 8. ๐‘ƒ ๐‘ฅาง > 12 = ๐‘ƒ ๐‘ > 12 โˆ’ 10 8ฮค 16 = ๐‘ƒ ๐‘ > 1 = 1 โˆ’ 0.8413 = 0.1587 MiamiOH.edu/cas @CASMiamiOH 9 Applying CLT with Sample Mean Example: The duration of Alzheimerโ€™s disease from the time symptoms first appear until death ranges from 3 to 20 years; the average is 8 years with standard deviation of 4 years.

The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimerโ€™s patients from the medical centerโ€™s database and records the duration of the disease. Find them approximate probabilities for these events; 1. The average duration is less than 7 years. 2. The average duration exceeds 7 years. 3. The average duration lies within 1 year of the population mean ๐œ‡ = 8. Since the administrator has selected a random sample of 30 files from the database we can use the CLT to draw conclusions about the population. MiamiOH.edu/cas @CASMiamiOH 10 Applying CLT with Sample Mean Example Continued: เดฅ: Approximately Normal with Mean ๐œ‡ = 8 and standard Sampling Distribution of ๐’™ deviation ๐œŽ ๐‘› = 4 30 = 0.73. This is ensured by CLT with a sample size of ๐‘› = 30.

1. The average duration is less than 7 years. ๐‘ง= ๐‘ฅาง โˆ’ ๐œ‡ 7 โˆ’ 8 = = โˆ’1.37 0.73 ๐œŽฮค ๐‘› Using ๐‘-score table, ๐‘ƒ ๐‘ฅาง < 7 = ๐‘ƒ ๐‘ < โˆ’1.37 = 0.0853 2. The average duration exceeds 7 years. Using the complement rule. ๐‘ƒ ๐‘ฅาง > 7 = 1 โˆ’ ๐‘ƒ ๐‘ฅาง โ‰ค 7 = 1 โˆ’ 0.0853 = 0.9147 3. The average duration lies within 1 year of the population mean ๐œ‡ = 8. ๐‘ง= ๐‘ฅาง โˆ’ ๐œ‡ 9 โˆ’ 8 = = 1.37 0.73 ฮค ๐œŽ ๐‘› Using ๐‘-score table and results from 1 ๐‘ƒ 7 < ๐‘ฅาง < 9 = ๐‘ƒ โˆ’1.37 < ๐‘ < 1.37 = 0.9147 โˆ’ 0.0853 = 0.8294 MiamiOH.edu/cas @CASMiamiOH 11 CLT and the Binomial Random Variable โ€ข The Central Limit Theorem can be used to conclude that the binomial random variable ๐‘‹ is approximately normal when ๐‘› is large, with mean ๐‘›๐‘ and standard deviation ๐‘›๐‘๐‘ž โ€ข The sample proportion, ๐‘ฦธ = ๐‘ฅ ๐‘› is simply a rescaling of the binomial random variable ๐‘‹, dividing it by ๐‘›. โ€ข From the Central Limit Theorem, the sampling distribution of ๐‘ฦธ will also be approximately normal, with a rescaled mean and standard deviation Remember to check the assumptions of the normal approximation of the binomial distribution. ๐‘›๐‘ > 5 and ๐‘›๐‘ž > 5 MiamiOH.edu/cas @CASMiamiOH 12 Sampling Distribution of the Sample Proportion โ€ข A random sample of size ๐‘› is selected from a population that follows the binomial distribution with parameter ๐‘.

โ€ข The sampling distribution of the sample proportion, ๐‘ฦธ = will have a mean ๐‘ and standard deviation โ€ข ๐‘ฅ ๐‘› ๐‘๐‘ž ๐‘› If ๐‘› is large, and ๐‘ is not too close to zero or one, the sampling distribution of ๐‘ฦธ will be approximately normal. The standard deviation of ๐‘ฦธ is sometimes called the Standard Error (SE). MiamiOH.edu/cas @CASMiamiOH 13 Probabilities for the Sample Proportion โ€ข If the sampling distribution of ๐‘ฦธ is normal or approximately normal, standardize or rescale the interval of interest in terms of ๐‘ง= โ€ข ๐‘ฦธ โˆ’ ๐‘ ๐‘๐‘ž ๐‘› Find the appropriate area using a Z-Score table We have, ๐‘†: Proportion of Underfilled cans ๐‘› = 200; ๐‘ = 0.05; ๐‘ž = 0.95 ๐‘›๐‘ = 10 > 5 and ๐‘›๐‘ž = 190 > 5 Passes Assumptions Example: The soda bottler claims that only 5% of the soda cans are underfilled. A quality control technician randomly 0.1โˆ’0.05 samples 200 cans. What is the probability ๐‘ƒ ๐‘ฦธ > 0.1 = ๐‘ƒ ๐‘ > 0.05(0.95)ฮค200 = ๐‘ƒ ๐‘ > 3.24 that more than 10% of the cans are = 1 โˆ’ 0.99994 = 0.0006 underfilled?

This would be very unusual, if indeed ๐‘ = 0.05! MiamiOH.edu/cas @CASMiamiOH 14 Applying CLT with Sample Proportion Example: A random sample of 500 parents were surveyed about the importance of sports for boys and girls. Of the parents interviewed 60% agreed that boys and girls should have equal opportunities to participate in sports. Suppose someone claims the true proportion of parents in the population is actually equal to 55%. Since the 500 parents were randomly selected and ๐‘›๐‘ = 500 โ‹… 0.55 = 275 โ‰ซ 5 and ๐‘›๐‘ž = 500 โ‹… 0.45 = 225 โ‰ซ 5, we can use the CLT to draw conclusions about the population.

What is the probability of observing a sample proportion as large as or larger than the observed value ๐‘ฦธ = 0.6? ๐‘ง= ๐‘ฦธ โˆ’ ๐‘ 0.6 โˆ’ 0.55 = = 2.25 0.0222 ฮค ๐‘๐‘ž ๐‘› Using ๐‘-score table we find ๐‘ƒ ๐‘ฦธ > 0.6 โ‰ˆ ๐‘ƒ ๐‘ > 2.25 = 1 โˆ’ 0.9878 = 0.0122 Notice an observation of 60% is a pretty rare event. MiamiOH.edu/cas @CASMiamiOH 15 Large Sample Estimation Introduction โ€ข Populations are described by their probability distribution and parameters. For quantitative populations, the location and shape are described by ๐œ‡ and ๐œŽ. โ€ข For binomial populations, the locations and shape are determined by ๐‘ โ€ข โ€ข If the values of parameters are unknown, we make inferences about them using sample information. MiamiOH.edu/cas @CASMiamiOH 16 Large Sample Estimation Types of Inferences โ€ข Estimation: โ€ข โ€ข โ€ข Estimating or predicting the value of the parameter.

What is/are the most likely values of ๐œ‡ or ๐‘? Hypothesis Testing: โ€ข Making decisions about the value of a parameter based on some preconceived idea. โ€ข Did the sample come from a population with ๐œ‡ = 5? Or similarly a population ๐‘ = 0.2? MiamiOH.edu/cas @CASMiamiOH 17 Large Sample Estimation Inferences Using the Estimation Using Hypothesis Testing Example: A consumer wants to estimate the average price of similar homes in their city before putting their house on the market. Example: A manufacturer want to know if a new type of steel is more resistant to high temperatures than the old type. Estimation: They estimate, ๐œ‡, the average home price by using the sample mean Hypothesis Test: Is the new average resistance, ๐œ‡new , equal to the old average resistance, ๐œ‡old ? MiamiOH.edu/cas @CASMiamiOH 18 Large Sample Estimation Types of Inferences โ€“ Continued โ€ข Whether you are estimating parameters or testing a hypothesis, statistical methods are important because they provide: โ€ข โ€ข Methods for making inferences A numerical measure of the goodness or reliability of that inference. MiamiOH.edu/cas @CASMiamiOH 19 Large Sample Estimation Estimators versus Estimation

โ€ข An estimator is a rule, usually a formula, that tells you how to calculate the estimate based on the sample. โ€ข Point Estimation: A single number calculated to estimate the parameter โ€ข Interval Estimation: Two numbers are calculated to create an interval within which the parameter is expected to lie. MiamiOH.edu/cas @CASMiamiOH 20 Large Sample Estimation Properties of Point Estimators โ€ข โ€ข โ€ข Since an estimator is calculated from sample values, it varies from sample to sample according to its sampling distribution. An estimator is unbiased if the mean of its sampling distribution equals the parameter of interest. Of all the unbiased estimators, the preferred estimators are those whose sampling distributions has the smallest spread or variability .

MiamiOH.edu/cas @CASMiamiOH 21 Large Sample Estimation Measuring the Goodness of an Estimator โ€ข The distance between an estimate and the true value of the parameter is the error of estimation. The distance between the arrow and the bullseye. โ€ข In this section, the sample sizes are large, so that our unbiased estimators will have normal distributions. โ€ข Recall: The Central Limit Theorem (CLT) MiamiOH.edu/cas @CASMiamiOH 22 Large Sample Estimation The Margin of Error โ€ข โ€ข For an unbiased estimator with a normal sampling distribution, 95% of all point estimates lie within 1.96 standard deviations of the parameter of interest. Margin of Error: The maximum error of estimation calculated as . Notice: The margin of error is 1.96 โ‹… SE of the estimator MiamiOH.edu/cas @CASMiamiOH 23 Large Sample Estimation Estimating Means and Proportions โ€ข For a quantitative population.

Point estimator of the population mean, ๐œ‡: ๐‘ฅาง ๐‘  Margin of Error (๐‘› โ‰ฅ 30): ยฑ1.96 ๐‘› โ€ข For a binomial population ๐‘ฅ Point estimator of the population proportion, ๐‘ฦธ : ๐‘› Margin of Error (๐‘› โ‰ฅ 30): ยฑ1.96 ๐‘เทœ๐‘žเทœ ๐‘› MiamiOH.edu/cas @CASMiamiOH 24 Large Sample Estimation Interval Estimation โ€ข โ€ข โ€ข Create an interval (๐‘Ž, ๐‘) so that you are fairly sure that the parameter lies between these two values. โ€œFairly sureโ€ means with high probability, as measured by the confidence coefficient. Usually, 1 โˆ’ ๐›ผ = 0.9, 0.95, 0.98 or 0.99 Suppose 1 โˆ’ ๐›ผ = 0.95 and that the estimator has a normal distribution. Estimator ยฑ1.96 โ‹… ๐‘†๐ธ MiamiOH.edu/cas @CASMiamiOH 25 Large Sample Estimation Interval Estimation โ€“ Continued

โ€ข Since we donโ€™t know the value of the parameter consider Estimator ยฑ 1.96 โ‹… SE โ€ข . Only if the estimate falls in the tail areas will the interval fail to enclose the parameter. This only happens 5% of the time. MiamiOH.edu/cas @CASMiamiOH 26 Large Sample Estimation Different Confidence Levels โ€ข To change a general confidence level , 1 โˆ’ ๐›ผ, pick a value of ๐‘ that puts area 1 โˆ’ ๐›ผ in the center of the ๐‘ distribution Tail Area ๐’ ๐œถ ฮค๐Ÿ 0.05 1.645 0.025 1.96 0.01 2.33 0.005 2.58 100 1 โˆ’ ๐›ผ % confidence interval: Estimator ยฑ ๐‘๐›ผฮค2 โ‹… SE . MiamiOH.edu/cas @CASMiamiOH 27 Large Sample Estimation Confidence Interval for Means and Proportions โ€ข For a quantitative population. Confidence interval for a population mean, ๐œ‡: ๐‘ฅาง ยฑ ๐‘๐›ผฮค2 โ€ข ๐‘  ๐‘› For a binomial population Confidence interval for a population proportion, ๐‘ : ๐‘ฦธ ยฑ ๐‘๐›ผฮค2 ๐‘ฦธ ๐‘žเทœ ๐‘› MiamiOH.edu/cas @CASMiamiOH 28 Large Sample Estimation Estimating the Difference Between Two Means โ€ข โ€ข Sometimes we are interested in comparing the means of two populations.

We define our random sample as follows: โ€ข โ€ข โ€ข Sample 1: a random sample of size ๐‘›1 drawn from population 1 with mean ๐œ‡1 and variance ๐œŽ12 Sample 2: a random sample of size ๐‘›2 drawn from population 2 with mean ๐œ‡2 and variance ๐œŽ22 We compare the two averages by making inferences about ๐œ‡1 โˆ’ ๐œ‡2 , the difference in the two population averages. โ€ข โ€ข If the two populations are the same, then ๐œ‡1 โˆ’ ๐œ‡2 = 0 The best estimate of ๐œ‡1 โˆ’ ๐œ‡2 is the difference in the two sample means , ๐‘ฅาง1 โˆ’ ๐‘ฅาง2 . MiamiOH.edu/cas @CASMiamiOH 29 Large Sample Estimation เดฅ๐Ÿ โˆ’ ๐’™ เดฅ๐Ÿ The Sampling Distribution of ๐’™ โ€ข The mean of ๐‘ฅาง1 โˆ’ ๐‘ฅาง2 is ๐œ‡1 โˆ’ ๐œ‡2 , the difference in the population means. ๐œŽ12 ๐‘›1 + ๐œŽ22 ๐‘›2 โ€ข The standard deviation of ๐‘ฅาง1 โˆ’ ๐‘ฅาง2 is SE = โ€ข If the sample sizes are large, the sampling distribution of ๐‘ฅาง1 โˆ’ ๐‘ฅาง2 is approximately normal, and SE can be estimated as SE = ๐‘ 12 ๐‘›1 + ๐‘ 22 . ๐‘›2 MiamiOH.edu/cas @CASMiamiOH 30 Large Sample Estimation Estimating ๐๐Ÿ โˆ’ ๐๐Ÿ โ€ข For large samples, point estimates and their margin of error as well as confidence intervals are based on the standard normal (๐‘) distribution. Point Estimate for ๐œ‡1 โˆ’ ๐œ‡2 : ๐‘ฅาง1 โˆ’ ๐‘ฅาง2 Margin of Error: ยฑ1.96 โ‹…

โ€ข ๐‘ 12 ๐‘›1 + ๐‘ 22 ๐‘›2 Confidence Interval for ๐œ‡1 โˆ’ ๐œ‡2 : (๐‘ฅาง1 โˆ’ ๐‘ฅาง2 ) ยฑ ๐‘๐›ผฮค2 โ‹… ๐‘ 12 ๐‘›1 + ๐‘ 22 ๐‘›2 The confidence interval contains the value ๐œ‡1 โˆ’ ๐œ‡2 = 0. Therefore, it is possible that ๐œ‡1 = ๐œ‡2 . You would not want to conclude that there is a difference in averages between the two populations. MiamiOH.edu/cas @CASMiamiOH 31 Large Sample Estimation เดฅ๐Ÿ โˆ’ ๐’™ เดฅ๐Ÿ , Example ๐’™ Tire 1 Tire 2 ๐‘ฅ1าง = 26,400 miles ๐‘ฅาง2 = 25,100 miles ๐‘ 12 = 1,440,000 ๐‘ 22 = 1,960,000 The wearing qualities of two types of automobile tires were compared by road-testing samples of ๐‘›1 = ๐‘›2 = 100 tires for each type and recording the number of miles until wearout, defined as a specific amount of tire wear. (Results given in table.) Estimate (๐œ‡1 โˆ’ ๐œ‡2 ), the difference in mean miles to wearout, using a 99% confidence interval. Is there a difference in the average wearing quality for the two types of tires? Computing the Point Estimate of (๐๐Ÿ โˆ’ ๐๐Ÿ ): ๐‘ฅ1าง = ๐‘ฅาง2 = 26,400 = 25,100 = 1,300 miles confidence interval we have, ๐‘ 2 ๐‘ 2 1,440,000 1,960,000 Standard Error of (เดฅ ๐’™๐Ÿ โˆ’ เดฅ ๐’™๐Ÿ ): 1 + 2 = + = 184.4 miles ๐‘›1 ๐‘›2 100 100 824.2 < ๐œ‡1 โˆ’ ๐œ‡2 < 1,775.8.

The difference in the average miles to wearout for the two types of tires is estimated to lie between the lower confidence limit 824.2 and upper confidence limit of 1,775.8. MiamiOH.edu/cas @CASMiamiOH 32 Large Sample Estimation Estimating the Difference Between Two Proportions โ€ข โ€ข Sometimes we are interested in comparing the proportion of โ€œsuccessesโ€ in two binomial populations. We define our random sample as follows: โ€ข โ€ข โ€ข Sample 1: a random sample of size ๐‘›1 drawn from binomial population 1 with parameter ๐‘1 Sample 2: a random sample of size ๐‘›2 drawn from binomial population 2 with parameter ๐‘2 We compare the two proportions by making inferences about ๐‘1 โˆ’ ๐‘2 , the difference in the two population proportions.

โ€ข โ€ข If the two populations are the same, then ๐‘1 โˆ’ ๐‘2 = 0 The best estimate of ๐‘1 โˆ’ ๐‘2 is the difference in the two sample proportions, ๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 = ๐‘ฅ1 ฮค๐‘›1 โˆ’ ๐‘ฅ2 ฮค๐‘›2 MiamiOH.edu/cas @CASMiamiOH 33 Large Sample Estimation เท๐Ÿ โˆ’ ๐’‘ เท๐Ÿ The Sampling Distribution of ๐’‘ โ€ข The mean of the sampling distribution of ๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 should be ๐‘1 โˆ’ ๐‘2 , as the difference in the population proportions. โ€ข The standard deviation of ๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 is SE = โ€ข If the sample sizes are large, the sampling distribution of ๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 is ๐‘1 ๐‘ž1 ๐‘›1 + ๐‘2 ๐‘ž2 ๐‘›2 approximately normal, and SE can be estimated as SE = ๐‘เทœ1 ๐‘žเทœ1 ๐‘›1 + MiamiOH.edu/cas ๐‘เทœ2 ๐‘žเทœ2 . ๐‘›2 @CASMiamiOH 34 Large Sample Estimation Estimating ๐’‘๐Ÿ โˆ’ ๐’‘๐Ÿ โ€ข For large samples, point estimates and their margin of error as well as confidence intervals are based on the standard normal (๐‘) distribution. Point Estimator for p1 โˆ’ ๐‘2 : ๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 Margin of Error: ยฑ1.96 โ‹… โ€ข ๐‘เทœ1 ๐‘žเทœ1 ๐‘›1 + ๐‘เทœ2 ๐‘žเทœ2 ๐‘›2 Confidence Interval for p1 โˆ’ ๐‘2 : ๐‘ฦธ1 ๐‘žเทœ1 ๐‘ฦธ 2 ๐‘žเทœ2 (๐‘ฦธ1 โˆ’ ๐‘ฦธ 2 ) ยฑ ๐‘๐›ผฮค2 โ‹… + ๐‘›1 ๐‘›2 The confidence interval contains the value p1 โˆ’ ๐‘2 = 0. Therefore, it is possible that p1 = ๐‘2 .

You would not want to conclude that there is a difference in proportions between the two populations. MiamiOH.edu/cas @CASMiamiOH 35 Developing Section Rest of City Sample Size, ๐‘› 50 100 Favoring 38 65 Large Sample Estimation เท๐Ÿ โˆ’ ๐ฉ เท๐Ÿ , Example ๐ฉ A bond proposal for school construction is on the P-hat 0.76 0.65 ballot at the next city election. Money from this bond issue will be used to build schools in rapidly developing section of the city, and the remainder will be used to renovate and update school buildings in the rest of the city. Data from the random sample of residents is given above. 1. Estimate the difference in the true proportions favoring the bond proposal with a 99% confidence interval. 2. If both samples were pooled into one sample of size ๐‘› = 150, with 103 in favor of the proposal, provide a point estimate of the proportion of city residents who will vote for the bond proposal. What is the margin of error? MiamiOH.edu/cas @CASMiamiOH 36 Developing Section Rest of City Sample Size, ๐‘› 50 100 Favoring 38 65 0.76 0.65 Large Sample Estimation เท๐Ÿ โˆ’ ๐ฉ เท๐Ÿ , Example Continued ๐ฉ Estimate the difference in the true proportions P-hat favoring the bond proposal with a 99% confidence interval. Point Estimate of ๐’‘๐Ÿ โˆ’ ๐’‘๐Ÿ : 0.76 โˆ’ 0.65 = 0.11 1. เท๐Ÿ ): Standard Error (เท ๐’‘๐Ÿ โˆ’ ๐’‘ ๐‘เทœ1 ๐‘žเทœ 2 ๐‘›1 + ๐‘เทœ2 ๐‘žเทœ2 ๐‘›2 = Co.

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