{"id":288,"date":"2021-07-14T15:59:07","date_gmt":"2021-07-14T15:59:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-population-means-with-known-standard-deviations\/"},"modified":"2023-12-05T09:38:46","modified_gmt":"2023-12-05T09:38:46","slug":"two-population-means-with-known-standard-deviations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-population-means-with-known-standard-deviations\/","title":{"raw":"More Testing for Two Population Means","rendered":"More Testing for Two Population Means"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Conduct a hypothesis test for a difference in two population means with known standard deviations and interpret the conclusion in context<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nEven though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is\u00a0[latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}[\/latex]. The normal distribution has the following format:\r\n\r\nNormal distribution<\/strong> is: [latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}\\sim{N}\\Bigg[{\\mu_{{1}}-\\mu_{{2}},\\sqrt{{\\dfrac{{(\\sigma_{{1}})}^{{2}}}{{n}_{{1}}}+\\dfrac{{(\\sigma_{{2}})}^{{2}}}{{n}_{{2}}}}}\\Bigg]}[\/latex]\r\n\r\nThe standard deviation<\/strong> is: [latex]\\displaystyle\\sqrt{\\dfrac{(\\sigma_1)^2}{n_1}+\\dfrac{(\\sigma_2)^2}{n_2}}[\/latex]\r\n\r\nThe\u00a0test statistic<\/strong> (z<\/em>-score) is: [latex]\\displaystyle{z}=\\dfrac{(\\overline{x}_1-\\overline{x}_2)-(\\mu_1-\\mu_2)}{\\sqrt{\\dfrac{(\\sigma_1)^2}{n_1}+\\dfrac{(\\sigma_2)^2}{n_2}}}[\/latex]\r\n
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    Recall: ORDER OF OPERATIONS<\/h3>\r\n
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    Please<\/strong><\/td>\r\nExcuse<\/strong><\/td>\r\nMy<\/strong><\/td>\r\nDear<\/strong><\/td>\r\nAunt<\/strong><\/td>\r\nSally<\/strong><\/td>\r\n<\/tr>\r\n
    parentheses<\/td>\r\nexponents<\/td>\r\nmultiplication<\/td>\r\ndivision<\/td>\r\naddition<\/td>\r\nsubtraction<\/td>\r\n<\/tr>\r\n
    [latex]( \\ )[\/latex]<\/td>\r\n[latex]x^2[\/latex]<\/td>\r\n[latex]\\times \\ \\mathrm{or} \\ \\div[\/latex]<\/td>\r\n[latex]+ \\ \\mathrm{or} \\ -[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    To calculate the test statistic (z-score) follow these steps:<\/h2>\r\nFirst, find the numerator. Calculate the difference between the two sample means [latex](\\overline{x}_1- \\overline{x}_2)[\/latex] and the two population means [latex](\\mu _1- \\mu _2)[\/latex], then subtract them.\u00a0<\/strong>\r\n\r\n \r\n\r\nSecond, find the denominator. You will end up taking the square root of the entire bottom; a square root can be understood as parentheses.\u00a0<\/strong>\r\n\r\nStep 1: Calculate [latex]\\frac{(\\sigma _1)^2}{n_1}[\/latex] by squaring the population standard deviation of the first population and then dividing by the sample size taken from the first population.\r\n\r\nStep 2: Calculate [latex]\\frac{(\\sigma _2)^2}{n_2}[\/latex] by squaring the population standard deviation of the second population and then dividing by the sample size taken from the second population.\r\n\r\nStep 3: Add the value you got in Step 1 and Step 2.\r\n\r\nStep 4: Find the square root of the value you found in Step 3.\r\n\r\nThird, take the numerator and divide by the denominator.<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Example 1<\/h3>\r\nIndependent groups, population standard deviations known.<\/strong> The mean lasting time of two competing floor waxes is to be compared. Twenty floors<\/strong> are randomly assigned to test each wax<\/strong>. Both populations have a normal distribution. The data are recorded in the table.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Wax<\/th>\r\nSample Mean Number of Months Floor Wax Lasts<\/th>\r\nPopulation Standard Deviation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1<\/td>\r\n3<\/td>\r\n0.33<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n2.9<\/td>\r\n0.36<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDoes the data indicate that\u00a0wax 1 is more effective than wax 2<\/strong>? Test at a 5% level of significance.\r\n\r\n[reveal-answer q=\"528070\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"528070\"]\r\n\r\nThis is a test of two independent groups, two population means, population standard deviations known.\r\n\r\nRandom Variable:\u00a0<\/strong>[latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}[\/latex] = difference in the mean number of months the competing floor waxes last.\r\n\r\nH0<\/sub><\/em>: \u03bc1<\/sub><\/em> \u2264 \u03bc<\/em>2<\/sub>\r\n\r\nHa<\/sub><\/em>: \u03bc1<\/sub><\/em> > \u03bc2<\/sub><\/em>\r\n\r\nThe words\u00a0\"is more effective\"<\/strong> say that wax 1 lasts longer than wax 2<\/strong>, on average. \"Longer\" is a \">\" symbol and goes into Ha<\/sub><\/em>. Therefore, this is a right-tailed test.\r\n\r\nDistribution for the test:<\/strong> The population standard deviations are known so the distribution is normal. Using the formula, the distribution is:\r\n\r\n[latex]{\\overline{x}_1}-{\\overline{x}_2}[\/latex] ~\u00a0N<\/em> [latex]\\left ( 0, {\\sqrt{{\\dfrac{0.33^2}{20}}+{\\dfrac{0.36^2}{20}}}} \\right )[\/latex]\r\n\r\nSince\u00a0\u03bc1<\/sub><\/em> \u2264 \u03bc2<\/sub><\/em> then \u03bc1<\/sub><\/em> \u2013 \u03bc2<\/sub><\/em> \u2264 0 and the mean for the normal distribution is zero.\r\n\r\nCalculate the p<\/em>-value using the normal distribution:<\/strong> p<\/em>-value = 0.1799\r\n\r\nGraph:<\/strong>\r\n\r\n\"This\r\n\r\n[latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}={3}-{2.9}={0.1}[\/latex]\r\n\r\nCompare \u03b1<\/em> and the p<\/em>-value:<\/strong> \u03b1<\/em> = 0.05 and p<\/em>-value = 0.1799. Therefore, \u03b1<\/em> < p<\/em>-value.\r\n\r\nMake a decision:<\/strong> Since \u03b1<\/em> < p<\/em>-value, do not reject H0<\/sub><\/em>.\r\n\r\nConclusion:<\/strong> At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.\r\n

    USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/h4>\r\n