answer:When simplifying sentences to improve clarity and concision in written communication, consider the following key factors: 1. **Remove unnecessary words and phrases**: Identify words or phrases that don't contribute to the sentence's core meaning and eliminate them. Examples include redundant expressions, like it is or there is, and filler words such as basically or essentially. 2. **Use active voice**: Sentences written in the active voice tend to be more concise and easier to understand than those in the passive voice. In active voice, the subject performs the action described by the verb, whereas in passive voice, the subject receives the action. For example, The manager wrote the report (active) is clearer than The report was written by the manager (passive). 3. **Avoid complex sentence structures**: Simplify sentences with multiple clauses or phrases by breaking them into shorter, more manageable sentences. This can improve the overall flow and make the text easier to follow. 4. **Use precise and concise vocabulary**: Choose words that accurately convey your intended meaning without using jargon or overly technical terms. Avoid using longer words or phrases when shorter alternatives are available. For instance, use is often more effective than utilize. 5. **Focus on one main idea per sentence**: Ensure that each sentence conveys a single key point or idea. Avoid including multiple ideas or concepts in a single sentence, as this can lead to confusion and decreased clarity. By focusing on one main idea per sentence, you can create a clear and concise narrative that effectively communicates your message.

answer:To construct a 95% confidence interval for the population mean, we'll use the t-distribution since the sample size is relatively small (n = 25). We'll use the following formula: x̄ - t * (s / √n) < μ < x̄ + t * (s / √n) where x̄ is the sample mean, t is the critical t-value, s is the sample standard deviation, n is the sample size, and μ is the population mean. First, we need to find the critical t-value. Since we're constructing a 95% confidence interval, the significance level (α) is 0.05, and the degrees of freedom (df) is n - 1 = 24. Using a t-table or calculator, we find the critical t-value to be approximately 2.064. Now we can plug in the values: x̄ = 27.6 t = 2.064 s = 5.59 n = 25 The margin of error is t * (s / √n) = 2.064 * (5.59 / √25) = 2.064 * (5.59 / 5) = 2.064 * 1.118 = 2.31. Now we can construct the confidence interval: 27.6 - 2.31 < μ < 27.6 + 2.31 25.29 < μ < 29.91 Therefore, the 95% confidence interval for the population mean is approximately (25.29, 29.91). This interval suggests that, with 95% confidence, the true population mean lies between 25.29 and 29.91.

answer:The critical t-value of 2.064 comes from the t-distribution, which is a probability distribution used for estimating population parameters when the sample size is relatively small. The t-distribution is characterized by the degrees of freedom (df), which is the number of independent observations used to calculate the sample standard deviation, minus 1. In this case, the sample size (n) is 25, so the degrees of freedom (df) is 25 - 1 = 24. The critical t-value of 2.064 is the value on the t-distribution curve that separates the middle 95% of the area under the curve from the outer 5% (2.5% on each tail). This value is also known as the 97.5th percentile of the t-distribution with 24 degrees of freedom, since 95% of the area under the curve lies to the left of the upper critical value and 2.5% lies to the right. Think of it this way: if you were to take many random samples of size 25 from a normal population and calculate the t-statistic for each sample, about 95% of the t-statistics would fall within the range of -2.064 to 2.064. The remaining 5% would fall outside this range, with 2.5% in each tail. The 2.064 value can be found in a t-table, which provides critical t-values for various degrees of freedom and confidence levels. You can also use a calculator or software package to find the critical t-value. For example, the R function `qt(0.975, df = 24)` returns the value 2.064.

answer:To construct a 99% confidence interval, we'll use the same formula: x̄ - t * (s / √n) < μ < x̄ + t * (s / √n) The only difference is that we need to find the critical t-value for a 99% confidence interval. Since the significance level (α) is now 0.01, we'll look up the t-value that corresponds to the top 0.5% of the t-distribution (since 99% of the area under the curve lies to the left of the upper critical value and 0.5% lies to the right). Using a t-table or calculator, we find that the critical t-value for a 99% confidence interval with 24 degrees of freedom (df = 24) is approximately 2.797. Now we can plug in the values: x̄ = 27.6 t = 2.797 s = 5.59 n = 25 The margin of error is t * (s / √n) = 2.797 * (5.59 / √25) = 2.797 * (5.59 / 5) = 2.797 * 1.118 = 3.13. Now we can construct the confidence interval: 27.6 - 3.13 < μ < 27.6 + 3.13 24.47 < μ < 30.73 Therefore, the 99% confidence interval for the population mean is approximately (24.47, 30.73). This interval suggests that, with 99% confidence, the true population mean lies between 24.47 and 30.73. Notice that the 99% confidence interval is wider than the 95% confidence interval we calculated earlier. This makes sense, since we're more confident (99% vs 95%) that the true population mean lies within this wider interval.