I need help in writing a web services that parse data and compute various statis

Question

I need help in writing a web services that parse data and compute various statistics. You will write an app that uses those web services. You will write a web service, DataParserService that has a method, ParseData that accepts a string of numeric data separated by exactly one space and returns an array of doubles. No error checking is required; you can assume the user entered only doubles. You will write a web service, DescStatsService that has a method, GetDescStats that accepts an array of doubles and returns a DescriptiveStatistics object. This object contains the sample size, the average, and the standard deviation. No error checking is required. You will write a web service, CIService that has a method, GetCI that accepts an array of doubles and returns a ConfidenceInterval object. This object contains the (95%) lower class limit (LCL) and upper class limit (UCL). This web service is required to use the DescStatsService web service above. No error checking is required. You know how to calculate the sample size and average. The (sample) standard deviation (s) is found from: s=?((?_(i=1)^n??x_i^2-n¯x^2 ?)/(n-1)) where : ?_(i=1)^n?x_i^2 =x_1^2+x_2^2+?+x_n^2 A 95% confidence interval is calculated from: LCL= x ?-1.96 s/?n UCL= x ?+1.96 s/?n Note: this formula assumes a “large” sample which in practice is usually considered to be a sample of 30 or more data values. Your app should certainly work with this much data, but for debugging purposes

Type in data separated by spaces 4.4 2.33 5.26 8 11.1 3.49 9 2.22 Calculate Data Parser Service Results: 4.40, 2.33, 5.26, 8.00, 11.10, 3.49, 9.00, 2.22, s Service Results: Descriptive statistic Sample size: 8 Average: 5.73 Standard Deviation: 3.29 Confidence Interval Service Results: LCL: 3.45, UCL 8.00

Explanation / Answer

Now, compare the sampling distribution of x to the population distribution. Notice that both distributions are approximately normal with mean 53.4 inches. However, the sampling distribution of x is not as spread out as the population distribution. We can calculate the standard deviation of x as follows: Sample Mean, x 1 53.00 2 52.25 3 52.75 4 52.75 5 53.25 Table 22.1. Sample means. Unit 22: Sampling Distributions | Student Guide | Page 3 x = n x = 1.8 inches 4 0.9 inch Next, we put what we have learned about the sampling distribution of the sample mean to use in the context of manufacturing circuit boards. Although the scene depicted in the video is one that you don’t see much anymore in the United States, we can still explore how statistics can be used to help control quality in manufacturing. A key part of the manufacturing process of circuit boards is when the components on the board are connected together by passing it through a bath of molten solder. After boards have passed through the soldering bath, an inspector randomly selects boards for a quality check. A score of 100 is the standard, but there is variation in the scores. The goal of the quality control process is to detect if this variation starts drifting out of the acceptable range, which would suggest that there is a problem with the soldering bath. Based on historical data collected when the soldering process was in control, the quality scores have a normal distribution with mean 100 and standard deviation 4. The inspector’s random sampling of boards consists of samples of size five. Hence, the sampling distribution of x is normal with a mean of 100 and standard deviation of 4 / 5 1.79 . The inspector uses this information to make an x control chart, a plot of the values of x against time. A normal curve showing the sampling distribution of x has been added to the side of the control chart. Recall from the 68-95-99.7% rule, that we expect 99.7% of the scores to be within three standard deviations of the mean. So, we have added control limits that are three standard deviations (3 × 1.79 or 5.37 units) on either side of the mean . A point outside either of the control limits is evidence that the process has become more variable, or that its mean has shifted – in other words, that it’s gone out of control. As soon as an inspector sees a point such as the one outside the upper control limit in (For more information on control charts, . Control chart with control limits. So far we’ve been looking at population distributions that follow a roughly normal curve. Next, we look at a distribution of lengths of calls coming into the Mayor’s 24 Hour Hotline call center in Boston, Massachusetts. Most calls are relatively brief but a few last a very long time. The shape of the call-length distribution is skewed to the right . Duration of calls to a call center. To gain insight into the sampling distribution of the sample mean, x , for samples of size 10, we randomly selected 40 samples of size 10 and made a histogram of the sample means. We repeated this process for samples of size 20 and then again for samples of size 60. The histograms of the sample means appear in Sampling Distributions with the population distribution Notice that the spread of all the sampling distributions is smaller than the spread of the population distribution. Furthermore, as the sample size n increases, the spread of the sampling distributions decreases and their shape becomes more symmetric. By the time n = 60, the sampling distribution appears approximately normally distributed. What we have uncovered here is one of the most powerful tools statisticians possess, called the Central Limit Theorem. This states that, regardless of the shape of the population, the sampling distribution of the sample mean will be approximately normal if the sample size is sufficiently large. It is because of the Central Limit Theorem that statisticians can generalize from sample data to the larger population. We will be seeing applications of the Central Limit Theorem in later units on confidence intervals and significance tests

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