document.write( "Question 1200934: A mobile phone company is concerned about the lifetime of phone batteries supplied by a new supplier. Based upon historical data this type of battery should last for 900 days with a standard deviation of 150 days. A recent randomly selected sample of 40 batteries was selected and the sample battery life was found to be 942 days. Is the sample battery life significantly different from 900 days (significance level 5%)?\r
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document.write( "Ans: (Ho is accepted.) Am I right? \n" );
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Algebra.Com's Answer #835159 by math_tutor2020(3817) ![]() You can put this solution on YOUR website! \n" ); document.write( "mu = population mean \n" ); document.write( "sigma = population standard deviation \n" ); document.write( "n = sample size \n" ); document.write( "xbar = sample mean\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hypothesis: \n" ); document.write( "Ho: mu = 900 \n" ); document.write( "Ha: mu =/= 900 \n" ); document.write( "The \"not equals\" sign in the alternative hypothesis tells us that we're doing a two-tailed test.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Other given info \n" ); document.write( "sigma = 150 \n" ); document.write( "n = 40 \n" ); document.write( "xbar = 942\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Compute the test statistic. \n" ); document.write( "z = (xbar - mu)/(sigma/sqrt(n)) \n" ); document.write( "z = (942 - 900)/(150/sqrt(40)) \n" ); document.write( "z = 1.7708754896943 \n" ); document.write( "z = 1.77 \n" ); document.write( "Notice I used the Z statistic instead of T statistic. \n" ); document.write( "This is valid for two reasons
\n" ); document.write( " ![]() \n" ); document.write( "Image Source \n" ); document.write( "https://towardsdatascience.com/introduction-tfrom-the-central-limit-theorem-to-the-z-and-t-distributions-66513defb175\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now use a table such as this one \n" ); document.write( "https://www.ztable.net/ \n" ); document.write( "to find that \n" ); document.write( "P(Z < 1.77) = 0.96164 \n" ); document.write( "Or you can use a stats calculator for better accuracy.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This then leads to \n" ); document.write( "P(Z > 1.77) = 1 - P(Z < 1.77) \n" ); document.write( "P(Z > 1.77) = 1 - 0.96164 \n" ); document.write( "P(Z > 1.77) = 0.03836 \n" ); document.write( "This is the approximate area under the standard normal curve to the right of z = 1.77\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Because we're doing a two-tailed test, we'll double that result \n" ); document.write( "2*0.03836 = 0.07672 \n" ); document.write( "This represents the p-value. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Rule: If the p-value is smaller than alpha, then we reject the null. \n" ); document.write( "A handy memorization phrase could be \"If the p-value is low, then the null must go\".\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since the p-value (0.07672) is NOT smaller than alpha = 0.05, we fail to reject the null.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, the null is \"accepted\". I put that in quotes because technically we haven't accepted it. Rather we just haven't found enough evidence to reject it, so we have no choice but to side with the null.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Your answer of \"Ho is accepted\" is correct. \n" ); document.write( "One interpretation would be The sample battery life is not significantly different from 900 days \n" ); document.write( "In other words,The average lifespan of a cellphone battery appears to be 900 days\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Side notes:
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