Student Organization At A University Is Interested In Estima
Student Organization At A University Is Interested In Estimating
A student organization at a university is interested in estimating the proportion of students in favor of showing movies biweekly instead of monthly. A simple random sample (SRS) of 423 students revealed that 180 are in favor of showing movies biweekly.
The tasks involve calculating the point estimate, margin of error, and constructing a confidence interval for the proportion. Additionally, related questions include determining the z-score for a specified confidence level, and interpreting confidence intervals in the context of population means and proportions.
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Paper For Above instruction
Introduction
Statistical inference is crucial for understanding population characteristics based on sample data. When a student organization seeks to estimate the proportion of students favoring biweekly movies, key steps include calculating the point estimate, margin of error, and confidence interval. These computations provide an understanding of the likely true proportion, considering sampling variability. This paper details the calculation of these statistics, the interpretation of the confidence interval, and related concepts involving confidence levels and population means, contextualized through examples.
Point Estimate of Population Proportion
The point estimate is the sample proportion, representing the best estimate of the population proportion based on sample data. Given a sample of 423 students where 180 favor biweekly movies, the point estimate (p̂) is calculated as:
Rounded to two decimal places, the point estimate is 0.43. This indicates that approximately 43% of the student population favors showing movies biweekly instead of monthly. The point estimate provides a straightforward summary but does not account for sampling variability.
Margin of Error at 90% Confidence Level
The margin of error assesses the extent of sampling variability and determines the range within which the true population proportion likely falls, with a specified confidence level. To calculate the margin of error (ME), use the formula:
ME = z* × √[p̂(1 - p̂) / n]
where:
- p̂ = 0.4253 (from earlier),
- n = 423 (sample size),
- z* = z-value corresponding to the 90% confidence level.
From standard z-tables, the z-value for a 90% confidence level is approximately 1.64 (since the middle 90% of the standard normal distribution corresponds to z = ±1.64).
Calculating the standard error (SE):
SE = √[0.4253 × (1 - 0.4253) / 423] ≈ √[0.4253 × 0.5747 / 423] ≈ √[0.2444 / 423] ≈ √0.0005779 ≈ 0.02405
Now, the margin of error:
ME ≈ 1.64 × 0.02405 ≈ 0.0394
Rounded to two decimal places, the margin of error is 0.04. Therefore, with 90% confidence, the true proportion of students in favor of biweekly movies is within:
- Point estimate ± margin of error: 0.43 ± 0.04,
which gives the interval [0.39, 0.47].
Constructing the 90% Confidence Interval
Using the point estimate and margin of error, the confidence interval is:
[0.43 - 0.04, 0.43 + 0.04] = [0.39, 0.47]
This interval means we are 90% confident that the true proportion of students favoring biweekly movies lies between 39% and 47%. It provides a range of values that likely contain the true proportion, accounting for sampling variability.
Related Questions: Z-value for Confidence Intervals
Another question involves finding the z-value for constructing a confidence interval for the mean number of tweets sent per month by high school children. Given:
- Sample size n = 200,
- Sample mean = 1,500 tweets,
- Population standard deviation σ = 250,
- Confidence level = 75%.
The z-value associated with a 75% confidence level corresponds to the critical value that captures the middle 75% of the standard normal distribution. Since the remaining 25% is split equally between the two tails, each tail has 12.5%. Using standard normal distribution tables or calculator:
z* ≈ 1.15
This is the absolute value of z separating the middle 75% from the outer 12.5% tails on each side.
For constructing the confidence interval, the critical z-value used is approximately 1.15, which aligns with the choices provided.
Interpreting Confidence Intervals for Population Means
Confidence intervals for the mean age of Walmart shoppers can be used for making inferences about the population. Suppose a confidence interval is given that suggests the average age is above 52 or below 52.
- If the interval does not include 52 and is entirely above it, then we can conclude with the confidence level that the average age exceeds 52 (True).
- Conversely, if the interval is entirely below 52, the conclusion is that the average age is less than 52 (True).
- If the interval is entirely above 44, it indicates the average age is more than 44 (True). If not, the appropriate conclusion would depend on whether 44 is within the interval.
These interpretations rely on understanding that confidence intervals provide a range of plausible values for the population mean based on sample data and a specified confidence level.
Conclusion
Estimating proportions and means using sample data involves computing point estimates, margins of error, and constructing confidence intervals. These statistical tools allow researchers and organizations to make informed inferences about populations while accounting for sampling variability. Correct interpretation of confidence intervals helps in decision-making, policy formulation, and understanding population characteristics. As demonstrated through the examples provided, accurate calculations and interpretations are essential skills in statistical analysis, emphasizing the importance of understanding critical z-values, the logic of confidence intervals, and the context within which these statistical tools are applied.
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