{"id":219627,"date":"2025-05-26T12:37:51","date_gmt":"2025-05-26T12:37:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219627"},"modified":"2025-05-26T12:37:53","modified_gmt":"2025-05-26T12:37:53","slug":"ming-a-quality-assurance-analyst-at-a-bottling-factory-wants-to-use-a-one-sample-z-interval-to-estimate-the-proportion-of-500-ml-bottles-that-are-underfilled-she-requires-the-ma","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/26\/ming-a-quality-assurance-analyst-at-a-bottling-factory-wants-to-use-a-one-sample-z-interval-to-estimate-the-proportion-of-500-ml-bottles-that-are-underfilled-she-requires-the-ma\/","title":{"rendered":"Ming, a quality assurance analyst at a bottling factory, wants to use a\u00a0one-sample z-interval\u00a0to estimate the proportion of\u00a0500 mL bottles that are underfilled. She requires the\u00a0margin of error to be no more than \u00b14%\u00a0at a\u00a090% confidence level."},"content":{"rendered":"\n

Ming, a quality assurance analyst at a bottling factory, wants to use a one-sample z-interval<\/strong> to estimate the proportion of 500 mL bottles that are underfilled<\/strong>. She requires the margin of error to be no more than \u00b14%<\/strong> at a 90% confidence level<\/strong>.<\/p>\n\n\n\n

Required:<\/strong>
Determine the smallest sample size<\/strong> needed to achieve the desired margin of error.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To determine the smallest sample size<\/strong> required for Ming\u2019s estimate, we use the formula for the margin of error in a one-sample z-interval for a population proportion:<\/p>\n\n\n\n

Formula:<\/strong><\/h3>\n\n\n\n

ME=z\u2217\u22c5p^(1\u2212p^)nME = z^* \\cdot \\sqrt{\\frac{\\hat{p}(1 – \\hat{p})}{n}}<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • ME<\/strong> is the margin of error<\/li>\n\n\n\n
  • z\u2217z^*<\/strong> is the critical value corresponding to the confidence level<\/li>\n\n\n\n
  • p^\\hat{p}<\/strong> is the estimated proportion (we use 0.5 for maximum variability when no prior estimate is available)<\/li>\n\n\n\n
  • n<\/strong> is the sample size<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-step Calculation:<\/strong><\/h3>\n\n\n\n
      \n
    1. Confidence Level = 90%<\/strong>, so the critical value z\u2217z^* \u2248 1.645<\/strong><\/li>\n\n\n\n
    2. Desired Margin of Error (ME) = 0.04<\/strong><\/li>\n\n\n\n
    3. Assume p^=0.5\\hat{p} = 0.5<\/strong> to ensure the maximum required sample size<\/li>\n<\/ol>\n\n\n\n

      0.04=1.645\u22c50.5(1\u22120.5)n0.04 = 1.645 \\cdot \\sqrt{\\frac{0.5(1 – 0.5)}{n}} 0.04=1.645\u22c50.25n0.04 = 1.645 \\cdot \\sqrt{\\frac{0.25}{n}} 0.041.645=0.25n\\frac{0.04}{1.645} = \\sqrt{\\frac{0.25}{n}} (0.041.645)2=0.25n\\left(\\frac{0.04}{1.645}\\right)^2 = \\frac{0.25}{n} n=0.25(0.041.645)2n = \\frac{0.25}{\\left(\\frac{0.04}{1.645}\\right)^2} n\u22480.25(0.0243)2\u22480.250.000591\u2248423n \\approx \\frac{0.25}{(0.0243)^2} \\approx \\frac{0.25}{0.000591} \\approx 423<\/p>\n\n\n\n

      \n\n\n\n

      Answer:<\/strong><\/h3>\n\n\n\n

      The smallest sample size needed is approximately 423.<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation<\/strong><\/h3>\n\n\n\n

      Ming aims to estimate the proportion of underfilled 500 mL bottles in her bottling factory with a margin of error no greater than \u00b14% and a confidence level of 90%. To achieve this precision and confidence, determining the correct sample size is crucial.<\/p>\n\n\n\n

      In estimating a population proportion, the one-sample z-interval is appropriate when the population is large, and the sampling distribution of the sample proportion is approximately normal. This is generally true when both npnp and n(1\u2212p)n(1-p) are at least 10. Since the true proportion is unknown, Ming must assume the worst-case scenario for variability\u2014this occurs when the sample proportion p^=0.5\\hat{p} = 0.5, which maximizes the product p^(1\u2212p^)\\hat{p}(1 – \\hat{p}).<\/p>\n\n\n\n

      The formula for the margin of error for a population proportion involves the critical value from the standard normal distribution, the estimated proportion, and the sample size. For a 90% confidence level, the critical value is approximately 1.645.<\/p>\n\n\n\n

      By rearranging the formula to solve for the sample size nn, we find that a sample of at least 423 bottles is required. This ensures that the estimate of the proportion of underfilled bottles will be within \u00b14% of the true proportion, 90% of the time. Using this approach helps ensure statistically valid and reliable quality control in the factory\u2019s production process.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Ming, a quality assurance analyst at a bottling factory, wants to use a one-sample z-interval to estimate the proportion of 500 mL bottles that are underfilled. She requires the margin of error to be no more than \u00b14% at a 90% confidence level. Required:Determine the smallest sample size needed to achieve the desired margin of error. The Correct Answer and Explanation is: To […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-219627","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219627","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=219627"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219627\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219627"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219627"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219627"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}