Points In A Random Sample Of 198 People: 170 Said They Watch

1 10 Pts In A Random Sample Of 198 People 170 Said That They Watc

In a survey conducted with a random sample of 198 people, where 170 reported watching educational television, the task is to compute the 92% confidence interval for the true proportion of people who watch educational television. Additionally, the scenario involves various statistical problems including sample size determination, hypothesis testing, and interpreting the results of multiple significance tests. These problems encompass concepts of confidence intervals, margin of error, sample size estimation, hypothesis testing using the P-value method, and applications of the binomial and normal distributions in real-world research contexts.

Paper For Above instruction

Understanding the parameters and results of statistical inference is crucial in accurately interpreting survey data and experimental outcomes. This paper discusses the calculation of confidence intervals for proportions, determination of sample sizes for estimating population proportions, hypothesis testing about means, and the behavior of repeated significance tests over time. It emphasizes the importance of these tools in research across fields like public opinion polling, environmental studies, healthcare, and education.

Confidence Interval for a Population Proportion

Given a sample size \( n = 198 \), with \( x = 170 \) people watching educational television, the sample proportion \( p̂ \) is computed as:

\[ p̂ = \frac{x}{n} = \frac{170}{198} \approx 0.8586 \]

To construct the 92% confidence interval for the true proportion \( p \), we use the formula:

\[ CI = p̂ \pm z_{\alpha/2} \times \sqrt{\frac{p̂(1 - p̂)}{n}} \]

where \( z_{\alpha/2} \) is the critical value from the standard normal distribution for the desired confidence level. For 92%, \( \alpha = 0.08 \), so \( z_{0.04} \) is approximately 1.75.

Calculating the standard error:

\[ SE = \sqrt{\frac{0.8586 \times (1 - 0.8586)}{198}} \approx \sqrt{\frac{0.8586 \times 0.1414}{198}} \approx 0.025 \]

Thus, the margin of error (ME) is:

\[ ME = 1.75 \times 0.025 \approx 0.044 \]

Therefore, the confidence interval is:

\[ (0.8586 - 0.044, \quad 0.8586 + 0.044) = (0.8146, \; 0.9026) \]

This interval suggests that between approximately 81.5% and 90.3% of the population watches educational television with 92% confidence.

Sample Size for Estimating Population Proportion

Determining the necessary sample size to estimate a population proportion with a specified margin of error is fundamental for efficient survey design. Using the formula:

\[ n = \left( \frac{z_{\alpha/2} \times \sqrt{p(1-p)}}{E} \right)^2 \]

where \( p \) is the anticipated proportion, \( E = 0.05 \), and \( z_{0.025} \approx 1.96 \) for 95% confidence. Since the true proportion is unknown, a conservative estimate uses \( p = 0.5 \):

\[ n = \left( \frac{1.96 \times \sqrt{0.5 \times 0.5}}{0.05} \right)^2 = \left( \frac{1.96 \times 0.5}{0.05} \right)^2 = (19.6)^2 \approx 384 \]

The required sample size is approximately 384 respondents to ensure the estimate of the population proportion within a 5% margin of error at 95% confidence level.

Hypothesis Testing on Population Mean with Known Standard Deviation

In testing whether the average wind speed exceeds a claimed 8 mph, a sample of 32 days yielded a mean of 8.2 mph with a population standard deviation of 0.6 mph. The hypotheses are:

\[

H_0: \mu = 8 \quad \text{vs} \quad H_a: \mu \neq 8

\]

The test statistic for known population standard deviation is:

\[

z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{8.2 - 8}{0.6 / \sqrt{32}} \approx \frac{0.2}{0.106} \approx 1.89

\]

Using the standard normal distribution, the P-value is:

\[

P = 2 \times P(Z > 1.89) \approx 2 \times 0.0294 = 0.0588

\]

Since \( P \approx 0.0588 \) exceeds the significance level \( \alpha = 0.05 \), there is insufficient evidence to reject \( H_0 \). Therefore, the data do not provide enough evidence to conclude that the true mean wind speed differs from 8 mph at the 0.05 significance level.

Hypothesis Testing on Mean Number of Infections

A hospital's claim of an average of 16.3 infections per week is tested against a sample mean of 17.7 infections from 10 weeks, with a standard deviation of 1.8, assuming normality. The hypotheses are:

\[

H_0: \mu = 16.3 \quad \text{vs} \quad H_a: \mu \neq 16.3

\]

Using the t-test because the standard deviation is from a small sample:

\[

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{17.7 - 16.3}{1.8 / \sqrt{10}} \approx \frac{1.4}{0.569} \approx 2.46

\]

The degrees of freedom are \( df = 9 \). The P-value based on the t-distribution with 9 df:

\[

P \approx 2 \times P(t_{9} > 2.46) \approx 2 \times 0.0217 = 0.0434

\]

Since \( P

Sample Size Estimation for Estimating a Mean

The goal is to determine how many students need to be sampled to estimate the average GPA within 0.21 points with 95% confidence, given a known population standard deviation of 1.3. Using:

\[

n = \left( \frac{z_{\alpha/2} \times \sigma}{E} \right)^2

\]

where \( z_{0.025} \approx 1.96 \), \( \sigma = 1.3 \), and \( E = 0.21 \):

\[

n = \left( \frac{1.96 \times 1.3}{0.21} \right)^2 \approx \left( \frac{2.548}{0.21} \right)^2 \approx (12.13)^2 \approx 147.2

\]

The required sample size is approximately 148 students to achieve the desired precision.

Multiple Significance Tests and Null Hypothesis Behavior

When conducting multiple hypothesis tests, understanding the probability of false positives (Type I errors) across multiple tests is critical. If the null hypothesis is true in every case, the number of rejections follows a binomial distribution with parameters \( n=100 \) and \( p=\alpha=0.05 \). The expected number of false rejections (Type I errors) under the null is \( np = 5 \). The probability of rejecting the null exactly \( k \) times is:

\[

P(X = k) = \binom{100}{k} (0.05)^k (0.95)^{100-k}

\]

In the scenario where she rejects in 7 cases, this outcome could reasonably occur under the null hypothesis, given the properties of the binomial distribution. Specifically, observing 7 rejections is within the tail probabilities of the distribution, though slightly above the expected value of 5, suggesting that such a result could arise due to random variation, and does not necessarily imply the null is false in all cases. Multiple testing adjustments (e.g., Bonferroni correction) should be considered for controlling overall error rates.

Conclusion

The application of statistical inference, including confidence intervals, sample size calculations, hypothesis testing, and understanding the distribution of multiple tests, plays a vital role in research data analysis. Accurate interpretation of these statistical tools ensures valid conclusions about population parameters, leading to better decision-making in various fields from social sciences to healthcare.

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