To construct a 95% confidence interval for the proportion of college students who prefer printed textbooks, we will follow these steps:
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Sample Size (n): The sample size is given as 50 college students.
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Point Estimate (p̂): The point estimate \( p̂ \) (sample proportion) is calculated from the number of students preferring printed textbooks. \[ p̂ = \frac{\text{number who prefer printed textbooks}}{\text{total sample size}} = \frac{38}{50} = 0.76 \]
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Critical Value (z*): For a 95% confidence interval, we need the critical value associated with the standard normal distribution. This value corresponds to the 97.5th percentile (since we have 2.5% in each tail for a two-tailed test).
From the standard normal distribution table, the critical z-value for a 95% confidence level is approximately: \[ z^* \approx 1.96 \]
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Standard Error (SE): The standard error of the sample proportion is calculated using the formula: \[ SE = \sqrt{\frac{p̂(1 - p̂)}{n}} = \sqrt{\frac{0.76(1 - 0.76)}{50}} = \sqrt{\frac{0.76 \times 0.24}{50}} = \sqrt{\frac{0.1824}{50}} = \sqrt{0.003648} \approx 0.0604 \]
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Margin of Error (ME): To calculate the margin of error, we use the formula: \[ ME = z^* \times SE = 1.96 \times 0.0604 \approx 0.1184 \]
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Confidence Interval: The confidence interval is calculated as: \[ \text{Lower Limit} = p̂ - ME = 0.76 - 0.1184 \approx 0.6416 \] \[ \text{Upper Limit} = p̂ + ME = 0.76 + 0.1184 \approx 0.8784 \] Thus, the 95% confidence interval for the population proportion is approximately: \[ (0.6416, 0.8784) \]
Conclusion:
The claim from the study is that 61% of the population of all college students prefer printed textbooks. This is equivalent to a proportion of 0.61.
The confidence interval we constructed \((0.6416, 0.8784)\) does not contradict the study’s claim since this interval does not include the value 0.61. In fact, it suggests that the proportion of college students who prefer printed textbooks might be significantly higher than the 61% stated in the study.
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