1.A simple random sample of 400 individuals provides 100 Yes responses.
a. What is the point estimate of the proportion of the population that would provide Yes
responses?
b. What is your estimate of the standard error of the proportion, ?
c. Compute the 95% confidence interval for the population proportion.32.
- A simple random sample of 800 elements generates a sample proportion .70.
a. Provide a 90% confidence interval for the population proportion.
b. Provide a 95% confidence interval for the population proportion. - In a survey, the planning value for the population proportion is p* .35. How large a
sample should be taken to provide a 95% confidence interval with a margin of error of .05? - At 95% confidence, how large a sample should be taken to obtain a margin of error of .03
for the estimation of a population proportion? Assume that past data are not available for
developing a planning value for p*.
The correct answer and explanation is:
Let’s go step by step to solve these statistical problems.
Problem 1: Sample Proportion and Confidence Interval
Given:
- Sample size n=400n = 400
- Yes responses x=100x = 100
- Sample proportion p^=xn=100400=0.25\hat{p} = \frac{x}{n} = \frac{100}{400} = 0.25
(a) Point Estimate of Population Proportion
The point estimate of the population proportion is 0.25.
(b) Standard Error of the Proportion
The standard error (SE) of the sample proportion is given by: SE=p^(1−p^)nSE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} SE=0.25(1−0.25)400=0.1875400=0.00046875≈0.0217SE = \sqrt{\frac{0.25(1-0.25)}{400}} = \sqrt{\frac{0.1875}{400}} = \sqrt{0.00046875} \approx 0.0217
(c) 95% Confidence Interval
The confidence interval is calculated as: CI=p^±Zα/2×SECI = \hat{p} \pm Z_{\alpha/2} \times SE
For a 95% confidence level, the critical value Zα/2=1.96Z_{\alpha/2} = 1.96. CI=0.25±(1.96×0.0217)CI = 0.25 \pm (1.96 \times 0.0217) CI=0.25±0.0425CI = 0.25 \pm 0.0425 CI=(0.2075,0.2925)CI = (0.2075, 0.2925)
So, the 95% confidence interval is (0.2075, 0.2925).
Problem 2: Confidence Intervals for a Given Sample Proportion
Given:
- Sample size n=800n = 800
- Sample proportion p^=0.70\hat{p} = 0.70
(a) 90% Confidence Interval
For a 90% confidence level, Zα/2=1.645Z_{\alpha/2} = 1.645. SE=0.70(1−0.70)800=0.21800=0.0002625≈0.0162SE = \sqrt{\frac{0.70(1-0.70)}{800}} = \sqrt{\frac{0.21}{800}} = \sqrt{0.0002625} \approx 0.0162 CI=0.70±(1.645×0.0162)CI = 0.70 \pm (1.645 \times 0.0162) CI=0.70±0.0267CI = 0.70 \pm 0.0267 CI=(0.6733,0.7267)CI = (0.6733, 0.7267)
(b) 95% Confidence Interval
For a 95% confidence level, Zα/2=1.96Z_{\alpha/2} = 1.96. CI=0.70±(1.96×0.0162)CI = 0.70 \pm (1.96 \times 0.0162) CI=0.70±0.0317CI = 0.70 \pm 0.0317 CI=(0.6683,0.7317)CI = (0.6683, 0.7317)
Problem 3: Required Sample Size for 95% Confidence Interval with ME=0.05ME = 0.05
Using the formula: n=Zα/22⋅p∗(1−p∗)ME2n = \frac{Z_{\alpha/2}^2 \cdot p^* (1 – p^*)}{ME^2}
For a 95% confidence level, Zα/2=1.96Z_{\alpha/2} = 1.96, given p∗=0.35p^* = 0.35 and ME=0.05ME = 0.05: n=(1.96)2×0.35×(1−0.35)(0.05)2n = \frac{(1.96)^2 \times 0.35 \times (1-0.35)}{(0.05)^2} n=3.8416×0.35×0.650.0025n = \frac{3.8416 \times 0.35 \times 0.65}{0.0025} n=0.87330.0025=349.32n = \frac{0.8733}{0.0025} = 349.32
Rounding up, the required sample size is 350.
Problem 4: Required Sample Size for 95% Confidence Interval with ME=0.03ME = 0.03
Since no prior estimate is available, we assume p∗=0.5p^* = 0.5 (which maximizes sample size). n=(1.96)2×0.5×0.5(0.03)2n = \frac{(1.96)^2 \times 0.5 \times 0.5}{(0.03)^2} n=3.8416×0.250.0009n = \frac{3.8416 \times 0.25}{0.0009} n=0.96040.0009=1067.1n = \frac{0.9604}{0.0009} = 1067.1
Rounding up, the required sample size is 1068.
Explanation
Estimating population proportions using sample data is crucial in statistics. In the first case, we found the point estimate (sample proportion) for a sample of 400 individuals, calculated the standard error, and determined the 95% confidence interval. The interval tells us the range where the true population proportion is likely to fall with 95% certainty.
Similarly, for the second problem, a larger sample of 800 elements provided a sample proportion of 0.70. Using statistical formulas, we computed the confidence intervals at both 90% and 95% levels. A higher confidence level leads to a wider interval, indicating more certainty but less precision.
Determining the appropriate sample size is vital for accurate estimations. In the third problem, we used a planning proportion of 0.35 to determine that 350 individuals are required for a 95% confidence interval with a margin of error of 0.05. When no prior proportion estimate is available, we assume p∗=0.5p^* = 0.5 to get the maximum possible sample size. In the fourth problem, this approach led us to conclude that 1,068 individuals are needed to achieve a 95% confidence interval with a margin of error of 0.03.
These calculations illustrate the balance between confidence level, margin of error, and sample size. A larger sample reduces error but increases cost and effort. In real-world surveys, statisticians use these methods to ensure reliable and cost-effective data collection.
Now, generating an image related to these statistical concepts. 🎨
Here’s an infographic explaining confidence intervals and sample size determination in statistics. Let me know if you need any modifications or explanations!
