Confidence Interval Estimation

Inferential Statistics Objective: To estimate population characteristics using sample information.
Point Estimate: A sample characteristic (e.g., sample mean) used to estimate the corresponding population characteristic.
Margin of Error: A range of values around the point estimate, within which the population value is estimated to fall.
Confidence Level: The degree of certainty associated with the estimate.
Interval Estimate: A range of values within which the population value is believed to fall. Also known as the confidence interval.
Example Scenario:
- Estimating the average amount spent on lunch by UGH students (population mean).
- Surveying 200 students (sample size).
- Sample reveals an average lunch spending of $12 (sample mean).
- Inferring that all students spend an average of $12 (point estimate).
- Applying a margin of error of +/- $3.
- Deriving a range from $9 to $15 (interval estimate).
- Expressing 95% confidence in the estimation (confidence level).

Point Estimate: A single value derived from the sample to estimate the population value (typically the sample mean).
Margin of Error: Measures the maximum expected difference between sample results and the actual population values.
Interval Estimate: A range of values used to estimate the population characteristic, determined by adding and subtracting the margin of error from the point estimate (confidence interval).
Confidence Level: The probability that the interval estimate contains the actual population mean (parameter).

Z-Distribution: Used when the population standard deviation is known, OR when the sample size is at least 30.
T-Distribution: Used when the population standard deviation is unknown AND the sample size is less than 30.
Formulas:
- Mean $\pm$ (z) $(\frac{SD}{\sqrt{n}})$
- Mean $\pm$ (t) $(\frac{SD}{\sqrt{n}})$

Confidence Level	Alpha Two Tail	Alpha One Tail	Critical Value
80%	20%	10%	1.28
90%	10%	5%	1.65
95%	5%	2.5%	1.96
98%	2%	1%	2.33
99%	1%	0.05%	2.58

Confidence Level	Alpha Two-Tail	Alpha One-Tail	Deg. of Freedom (n - 1)	Critical Value
80%	20%	10%	1	3.078
90%	10%	5%	1	6.314
95%	5%	2.5%	1	12.706
98%	2%	1%	1	31.821
99%	1%	0.5%	1	63.657
80%	20%	10%	2	1.886
90%	10%	5%	2	2.920
95%	5%	2.5%	2	4.303
98%	2%	1%	2	6.965
99%	1%	0.5%	2	9.925
80%	20%	10%	3	1.638
90%	10%	5%	3	2.353
95%	5%	2.5%	3	3.182
98%	2%	1%	3	4.541
99%	1%	0.5%	3	5.841
80%	20%	10%	4	1.533
90%	10%	5%	4	2.132
95%	5%	2.5%	4	2.776
98%	2%	1%	4	3.747
99%	1%	0.5%	4	4.604

The table continues for degrees of freedom 5 through 10.

Problem: Estimating the average hourly fee for counseling services in the city.
Given:
- Sample Mean = $60
- Standard Deviation = $12
- Sample Size = 50
- Confidence Level = 95% (Z value = 1.96)
Solution:
- (a) Point Estimate: $60 (sample mean)
- (b) Margin of Error:
  - (\frac{(1.96)(12)}{\sqrt{50})} = $3.33
- (c) Interval Estimate:
  - Lower = 60 – 3.33 = $56.67
  - Upper = 60 + 3.33 = $63.33
- (d) Interpretation: We are 95% confident that the true average hourly wage for all counselors in the city lies between $56.67 and $63.33.

Problem: Estimating the average time it takes the police to respond to emergency calls.
Given:
- Sample Mean = 450 seconds
- Standard Deviation = 60 seconds
- Sample Size = 120
- Confidence Level = 90% (Z value = 1.65)
Solution:
- (a) Point Estimate: 450 seconds
- (b) Margin of Error:
  - $(\frac{(1.65)(60)}{\sqrt{120})} = 9.04$
- (c) Interval Estimate:
  - Lower = 450 – 9.04 = 440.96 seconds
  - Upper = 450 + 9.04 = 459.04 seconds
- (d) Interpretation: We are 90% confident that the true average response time for all emergency calls lies between 440.96 and 459.04 seconds.

Increasing the sample size from 120 to 480 (4 times increase):
- Margin of Error decreases.
Example:
- Sample Size = 120, Margin of Error = 9.04, Interval Estimate = 440.96 and 459.04
- Sample Size = 480, Margin of Error = 4.52, Interval Estimate = 445.48 and 454.52
When the sample size increases four times its original size, the margin of error will reduce by half.
Conversely, when the sample size is reduced by a quarter of its original size, then the margin of error will become twice as large.

Increasing the confidence level:
- The z-value becomes larger.
- The margin of error becomes larger.
- The interval estimate becomes wider.
- Estimate becomes less precise.
Z-values for different confidence levels:
- 90% (z-value = $\pm$ 1.65)
- 95% (z-value = $\pm$ 1.96)
- 99% (z-value = $\pm$ 2.58)
Need a delicate balance between confidence and precision.
All being equal:
- An increase in confidence level or an increase in standard deviation will increase the margin of error.
- However, an increase in sample size will decrease the margin of error.

Problem: What effect does increasing the confidence level from 90% to 99% have on the margin of error?
Given:
- Sample Mean = 450
- Standard Deviation = 60
- Sample Size = 480
- Confidence Level = 99% (Z value = 2.58)
Solution:
- (a) Point Estimate: 450
- (b) Margin of Error:
  - $(\frac{(2.58)(60)}{\sqrt{480})} = 7.07$
- (c) Interval Estimate:
  - Lower = 450 – 7.07 = 442.93
  - Upper = 450 + 7.07 = 457.07
- (d) Interpretation: We are 99% confident that the true average response time for all emergency calls will lie between 442.93 and 457.07 seconds.

When the population standard deviation is unknown and the sample size is less than 30, the T-distribution is used.
The t-distribution is influenced by the degree of freedom (n – 1).
T-critical value is obtained from t-distribution table under various degrees of freedom and confidence levels.
The t-distribution resembles the normal distribution but is more spread out, and the height of the curve is lower.
As the sample size increases, the t-distribution approaches the standard normal z-distribution. The approximation is quite close when the sample size is 30 or more.

Problem: Estimating the average salary of social workers in a town.
Given:
- Sample Mean = $45,500
- Standard Deviation = $12,500
- Sample Size = 20
- Confidence Level = 90%
- Degrees of Freedom = 20 – 1 = 19 (T value = 1.729)
Solution:
- (a) Point Estimate: $45,500
- (b) Margin of Error:
  - $(\frac{(1.729)(12500)}{\sqrt{20})} = 4832.70$
- (c) Interval Estimate:
  - Lower = 45500 – 4832.70 = $40,667.30
  - Upper = 45500 + 4832.70 = $50,332.70
- (d) Interpretation: We are 90% confident that the true average salary for all social workers in the entire town will lie between $40,667.30 and $50,332.70.

Problem: Estimating the average amount spent on food by households in a community.
Given:
- Sample Mean = $520
- Standard Deviation = $45
- Sample Size = 24
- Confidence Level = 95%
- Degrees of Freedom = 24 – 1 = 23 (T value = 2.069)
Solution:
- (a) Point Estimate: $520
- (b) Margin of Error:
  - (\frac{(2.069)(45)}{\sqrt{24})} = $19
- (c) Interval Estimate:
  - Lower = 520 – 19 = $501
  - Upper = 520 + 19 = $539
- (d) Interpretation: We are 95% confident that the true average amount that households will spend each month on food will lie between $501 and $539.