document.write( "Question 1206223: 1. A sample of 65 observations select from a somewhat normal population gave a mean of 2.67 with standard deviation of 0.75.
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document.write( "A sample of 50 observations was independently taken from normal distribution and the mean was 2.59 with standard deviation of 0.66
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document.write( "A. If the objective is to make a comparison of the mean of the two populations, which method is
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document.write( "appropriate and why?
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document.write( "B. Test whether the first population has greater mean as compared to the second. What is your conclusion at 5% level of significance? \n" );
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Algebra.Com's Answer #843582 by Theo(13342)![]() ![]() You can put this solution on YOUR website! sample 1 has 65 observations with a mean of 2.67 and a standard deviation of .75. \n" ); document.write( "sample 2 has 50 observations with a mean of 2.59 and a standard deviation of .66.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "it looks like a two sample t-test would be appropriate for this.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "test is to see if sample 1 mean is greater than sample 2 mean. \n" ); document.write( "level of significance is .05. \n" ); document.write( "it is one tail with the level of significant on the right side of the confidence interval.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i used the two sample t-test calculator at https://www.statskingdom.com/140MeanT2eq.html\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "inputs and results of the analysis are shown below.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "inuts:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "results:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at .05 right tail level of significance, critical t-score with 113 degrees of freedom would be equal to 1.66.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "right tail critical p-value is .05.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the test t-score was .597 \n" ); document.write( "the test p-value was .2758\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "since the test t-score was less than the critical t-score, the results were not significant. \n" ); document.write( "likewise, since the test p-value was greater than the critical p-value, the results were not significant. \n" ); document.write( "these results are always consistent with each other, i.e. both significant or both not significant conclusions.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the graph shows that the test t-score are clearly within the confidence interval.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the conclusion is that there is not sufficient information to conclude that the first population mean was greater than the second population mean. \n" ); document.write( "the population means are considered to be the same with differences due to normally expected variation in sample means.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the two sample t-test was used instead of the two sample z-test because the standard deviations were taken fom the samples and not from the population.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |