document.write( "Question 1199646: A psycho-linguist has been studying the frequency of text-speech (e.g., R U Ok?, LOL ) has been progressing into EMAIL correspondence. He collects a sample of 121 EMAIL messages and computes the mean (average) number of examples of text-speech per message. He also has a huge library of archived EMAIL messages from the year 2001. From these he is able to calculate the population average and standard deviation. Compute a inferential statistic to evaluate the hypothesis that the use of text-speech has increased in the years since 2001. Use an alpha level that would insure a maximum of a %5 risk of rejecting a true null hypothesis. \r
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document.write( "These are the obtained values:\r
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document.write( "Population Mean from 2001: 1.6 \r
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document.write( "Population Standard deviation from 2001: 1.25 \r
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document.write( "Sample mean from this year: 1.98 \r
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document.write( "Calculate and properly report your calculated statistic, and interpret your conclusion in one or two sentences. \n" );
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Algebra.Com's Answer #848168 by textot(100)![]() ![]() ![]() You can put this solution on YOUR website! **1. Set up Hypotheses**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H0):** μ = 1.6 (The mean number of text-speech instances in current emails is the same as in 2001) \n" ); document.write( "* **Alternative Hypothesis (H1):** μ > 1.6 (The mean number of text-speech instances in current emails is greater than in 2001)\r \n" ); document.write( "\n" ); document.write( "**2. Calculate the Test Statistic (z-score)**\r \n" ); document.write( "\n" ); document.write( "* Since we know the population standard deviation (σ = 1.25), we can use the z-test: \n" ); document.write( " * z = (x̄ - μ) / (σ / √n) \n" ); document.write( " * where: \n" ); document.write( " * x̄ = Sample mean (1.98) \n" ); document.write( " * μ = Population mean (1.6) \n" ); document.write( " * σ = Population standard deviation (1.25) \n" ); document.write( " * n = Sample size (121)\r \n" ); document.write( "\n" ); document.write( " * z = (1.98 - 1.6) / (1.25 / √121) \n" ); document.write( " * z = 0.38 / (1.25 / 11) \n" ); document.write( " * z = 0.38 / 0.1136 \n" ); document.write( " * z ≈ 3.34\r \n" ); document.write( "\n" ); document.write( "**3. Determine Critical Value**\r \n" ); document.write( "\n" ); document.write( "* **Significance Level (α) = 0.05** \n" ); document.write( "* This is a one-tailed test (since H1 is μ > 1.6). \n" ); document.write( "* Find the critical z-value for α = 0.05 in a standard normal distribution table. \n" ); document.write( "* The critical z-value is approximately 1.645.\r \n" ); document.write( "\n" ); document.write( "**4. Make a Decision**\r \n" ); document.write( "\n" ); document.write( "* **Calculated z-score (3.34) is greater than the critical value (1.645).** \n" ); document.write( "* **Therefore, we reject the null hypothesis.**\r \n" ); document.write( "\n" ); document.write( "**5. Conclusion**\r \n" ); document.write( "\n" ); document.write( "* At the 0.05 significance level, there is sufficient evidence to conclude that the mean number of text-speech instances in current emails is significantly higher than the mean number of text-speech instances in emails from 2001.\r \n" ); document.write( "\n" ); document.write( "**In summary:**\r \n" ); document.write( "\n" ); document.write( "* The calculated z-score is 3.34. \n" ); document.write( "* We reject the null hypothesis at the 0.05 significance level. \n" ); document.write( "* This suggests that the use of text-speech in emails has significantly increased since 2001. \n" ); document.write( " \n" ); document.write( " |