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.
These are the obtained values:
Population Mean from 2001: 1.6
Population Standard deviation from 2001: 1.25
Sample mean from this year: 1.98
Calculate and properly report your calculated statistic, and interpret your conclusion in one or two sentences.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Set up Hypotheses**
* **Null Hypothesis (H0):** μ = 1.6 (The mean number of text-speech instances in current emails is the same as in 2001)
* **Alternative Hypothesis (H1):** μ > 1.6 (The mean number of text-speech instances in current emails is greater than in 2001)
**2. Calculate the Test Statistic (z-score)**
* Since we know the population standard deviation (σ = 1.25), we can use the z-test:
* z = (x̄ - μ) / (σ / √n)
* where:
* x̄ = Sample mean (1.98)
* μ = Population mean (1.6)
* σ = Population standard deviation (1.25)
* n = Sample size (121)
* z = (1.98 - 1.6) / (1.25 / √121)
* z = 0.38 / (1.25 / 11)
* z = 0.38 / 0.1136
* z ≈ 3.34
**3. Determine Critical Value**
* **Significance Level (α) = 0.05**
* This is a one-tailed test (since H1 is μ > 1.6).
* Find the critical z-value for α = 0.05 in a standard normal distribution table.
* The critical z-value is approximately 1.645.
**4. Make a Decision**
* **Calculated z-score (3.34) is greater than the critical value (1.645).**
* **Therefore, we reject the null hypothesis.**
**5. Conclusion**
* 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.
**In summary:**
* The calculated z-score is 3.34.
* We reject the null hypothesis at the 0.05 significance level.
* This suggests that the use of text-speech in emails has significantly increased since 2001.
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