The range for a set of data is estimated to be 24. a. What is the planning value for the population standard deviation? b. At 95% confidence, how large a sample would provide a margin of error of 3? c. At 95% confidence, how large a sample would provide a margin of error of 2?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To answer these questions, we\u2019ll use statistical formulas for estimating the population standard deviation<\/strong> from the range<\/strong>, and then apply those to the margin of error formula<\/strong> for estimating sample size.<\/p>\n\n\n\n We estimate the standard deviation (\u03c3) from the range (R) using the range rule of thumb<\/strong>: \u03c3\u2248Range4\\sigma \\approx \\frac{\\text{Range}}{4}<\/p>\n\n\n\n Given that the range is 24: \u03c3\u2248244=6\\sigma \\approx \\frac{24}{4} = 6<\/p>\n\n\n\n \u2705 Answer<\/strong>: The planning value for the population standard deviation is 6<\/strong>.<\/p>\n\n\n\n To find the required sample size<\/strong> for a given margin of error (E)<\/strong> at 95% confidence<\/strong>, use the formula: n=(z\u22c5\u03c3E)2n = \\left(\\frac{z \\cdot \\sigma}{E}\\right)^2<\/p>\n\n\n\n Where:<\/p>\n\n\n\n n=(1.96\u22c563)2=(11.763)2=(3.92)2\u224815.37n = \\left(\\frac{1.96 \\cdot 6}{3}\\right)^2 = \\left(\\frac{11.76}{3}\\right)^2 = (3.92)^2 \\approx 15.37<\/p>\n\n\n\n Always round up<\/strong> for sample size: Same formula, but with E=2E = 2: n=(1.96\u22c562)2=(11.762)2=(5.88)2\u224834.57n = \\left(\\frac{1.96 \\cdot 6}{2}\\right)^2 = \\left(\\frac{11.76}{2}\\right)^2 = (5.88)^2 \\approx 34.57<\/p>\n\n\n\n Round up: In statistics, the range rule of thumb<\/strong> provides a simple estimate of the population standard deviation. By dividing the range by 4, you approximate the typical spread of data, assuming a relatively normal distribution. This is useful when only limited data is available, such as an estimated range.<\/p>\n\n\n\n Once the standard deviation is estimated, we can determine the necessary sample size to achieve a desired margin of error (E)<\/strong> for a population mean. The margin of error represents how close a sample mean is expected to be to the true population mean, with a certain level of confidence.<\/p>\n\n\n\n Using the formula: n=(z\u22c5\u03c3E)2n = \\left(\\frac{z \\cdot \\sigma}{E}\\right)^2<\/p>\n\n\n\n we plug in the z-score for 95% confidence, which is 1.96<\/strong>, a standard value from the normal distribution.<\/p>\n\n\n\n For a margin of error of 3, we need fewer observations because we’re willing to tolerate more error. But for a smaller error of 2, a larger sample is required to tighten the estimate around the true mean.<\/p>\n\n\n\n This process is a fundamental part of sample planning<\/strong>, ensuring that studies or surveys are neither underpowered (too few participants) nor unnecessarily large (wasting resources).<\/p>\n","protected":false},"excerpt":{"rendered":" The range for a set of data is estimated to be 24. a. What is the planning value for the population standard deviation? b. At 95% confidence, how large a sample would provide a margin of error of 3? c. At 95% confidence, how large a sample would provide a margin of error of 2? […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-24475","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24475","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=24475"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24475\/revisions"}],"predecessor-version":[{"id":24476,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24475\/revisions\/24476"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=24475"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=24475"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=24475"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\na. What is the planning value for the population standard deviation?<\/strong><\/h3>\n\n\n\n
\n\n\n\nb. At 95% confidence, how large a sample would provide a margin of error of 3?<\/strong><\/h3>\n\n\n\n
\n
\u2705 Answer<\/strong>: A sample size of 16<\/strong> is needed.<\/p>\n\n\n\n
\n\n\n\nc. At 95% confidence, how large a sample would provide a margin of error of 2?<\/strong><\/h3>\n\n\n\n
\u2705 Answer<\/strong>: A sample size of 35<\/strong> is needed.<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n