STAT 124 - Analysis of a Single Sample

Introduction to Statistics

Course Information

  • Course Code: STAT 124

  • Institution: University of British Columbia (UBC)

  • Focus: Analysis of a Single Sample

Analysis of One Sample - Example

Problem Statement

  • Sample Size: 200 university students.

  • Observations: 31 students are business majors, 169 others are not.

Data Summary

Observed Data

  • Total Sample Size ($n$): 200

  • Business Majors: 31

  • Other Majors: 169

  • Sample Proportion ($\, \, ar{p}$): pˉ=rac31200=0.155\bar{p} = rac{31}{200} = 0.155 or 15.5%

Standard Error and Margin of Error

Calculation of Standard Error (SE)

  • Standard Error Formula: SE=extsqrtracpˉ(1pˉ)nSE = ext{sqrt} rac{\bar{p}(1 - \bar{p})}{n}

  • Calculation:

    • SE=extsqrtrac0.155imes0.845200ext0.0256SE = ext{sqrt} rac{0.155 imes 0.845}{200} ext{≈} 0.0256

Margin of Error (ME) for 95% Confidence Level

  • Critical Value ($z$): 1.96

  • Margin of Error Formula: ME=zimesSEME = z imes SE

  • Calculation:

    • ME=1.96imes0.0256ext0.0501ME = 1.96 imes 0.0256 ext{≈} 0.0501

Confidence Interval Interpretation

Calculation of 95% Confidence Interval

  • Confidence Interval Formula: CI=pˉext±MECI = \bar{p} ext{±} ME

  • Calculation:

    • CI=0.155ext±0.050ext[0.105,0.205]CI = 0.155 ext{±} 0.050 ext{→} [0.105, 0.205]

Interpretation

  • We are 95% confident that the true proportion of business majors in the entire university population lies between 10.5% and 20.5%.

Defining the Standard Error

Conceptual Definition

  • Standard Error (SE) measures the dispersion of sample proportions around the true population proportion.

  • Significance:

    • Quantifies sampling error.

    • Indicates how much we expect our sample proportion ($ar{p}$) to miss the true population proportion ($p$).

    • Represents the standard deviation of the sampling distribution.

Mathematical Formula for Standard Error

Formula Overview

  • Standard Error Formula: SE=extsqrtracp(1p)nSE = ext{sqrt} rac{p(1-p)}{n}

  • Components:

    • $p$: Proportion of successes.

    • $1 - p$: Proportion of failures.

    • $n$: Sample size (number of observations).

The Law of Large Numbers

Importance of Sample Size

  • Larger sample sizes lead to decreased Standard Error because $n$ is in the denominator.

  • Example: To halve the Standard Error, you must quadruple the sample size.

Precision vs. Accuracy

  • A small Standard Error indicates high precision, suggesting that multiple samples will yield similar results.

The Concept of Interval Estimation

Definitions

  • Point Estimate: A single value (e.g., 51%) used to estimate a population parameter.

  • Problem with Point Estimates: Samples vary, making point estimates likely to be slightly inaccurate.

  • Solution: Use an Interval Estimate to provide a range of plausible values.

  • Definition of Confidence Interval: A range derived from sample data likely to contain the unknown population parameter.

Constructing the Confidence Interval

Construction Methodology

  • The Confidence Interval ($CI$) is constructed by adding and subtracting a Margin of Error from the sample proportion ($ar{p}$): CI=pˉext±zSECI = \bar{p} ext{±} z^* SE

Trade-offs in Construction

  • Confidence Level: Increasing the confidence level (e.g., from 95% to 99%) results in a wider interval.

  • Sample Size: Increasing $n$ narrows the interval (greater precision).

Correct Interpretation of Confidence Intervals

Understanding 95% Confidence

  • Interpretation: “95% of all possible samples of this size will produce an interval that captures the true population proportion.”

Common Misconceptions

  • Incorrect Interpretation: “There is a 95% probability the true proportion is between A and B.” (The true proportion is fixed, not a variable)

  • Correct Interpretation: “We used a process that is successful 95% of the time.”

Hypothesis Testing Approach

Application of Hypothesis Testing

  • This data can be utilized to test claims about the population.

  • Null Hypothesis ($H0$): $p = p0$ (e.g.,

    • The university claims 20% of students are business majors).

  • Alternative Hypothesis ($Ha$): $p eq p0$.

The Research Question

Claim vs. Observed Data

  • University Claim: 20% business majors ($p_0 = 0.20$).

  • Observed: In our sample of $n = 200$, 15.5% ($ar{p} = 0.155$).

Hypotheses Formulation
  • Null Hypothesis ($H_0$): $p = 0.20$

  • Alternative Hypothesis ($H_a$): $p
    eq 0.20$

Calculating the Z-Statistic

Formula and Calculation Steps

  • Z-Statistic Formula: z=racpˉp<em>0extSE</em>0z = rac{\bar{p} - p<em>0}{ ext{SE}</em>{0}}

  • Steps:

    1. Difference calculation: 0.1550.20=0.0450.155 - 0.20 = -0.045

    2. Standard Error calculation:

    • extSE0=extsqrtrac0.20imes0.802000.0283ext{SE}_{0} = ext{sqrt} rac{0.20 imes 0.80}{200} ≈ 0.0283

    1. Compute $z$: z=rac0.0450.0283=1.59z = rac{-0.045}{0.0283} = -1.59

Statistical Decision

Results Overview

  • Calculated $z$: -1.59

  • Critical $z$ (at $ ext{α} = 0.05$): ±1.96

Conclusion

  • Decision: Fail to Reject $H_0$

  • Interpretation: Since $|-1.59| < 1.96$, the result is not statistically significant. There is insufficient evidence to claim the actual proportion differs from 20%.

Practical Significance vs. Statistical Significance

Insights

  • Despite the result not being statistically significant, a 4.5% gap might be important for university planning.

  • Increasing Power: If the same 15.5% result comes from a sample of $n = 500$, the resulting $z$ would be -2.51, which is significant ($p < 0.05$).

A More Complicated Situation - Analysis of Preferences

Example Scenario

  • Random sample of 100 people asked about their preference between Starbucks and Tim Horton’s.

  • Responses:

    • Starbucks: 51

    • Tim Horton’s: 39

    • No Preference: 10

Inquiry
  • Question posed: How to analyze these data?

Study Overview and Data Distribution

Sample Details

  • Random Sample Size ($n$): 100

  • Population: Large general population

Summary of Survey Responses

  • Starbucks: 51 (51%)

  • Tim Horton’s: 39 (39%)

  • No Preference: 10 (10%)

  • Total: 100 (100%)

Methodological Approaches for Analysis

Approaches to Analyze Categorical Frequency Data

  1. Chi-Square Goodness-of-Fit Test:

    • Tests if distribution across categories differs from theoretical expectation (e.g., an even distribution).

  2. Z-Test for Proportions:

    • Used for direct comparison between two specific groups (e.g., Starbucks vs. Tim Horton’s), with the third group as an external variable.

Precision: The Confidence Interval Analysis

Calculation of Confidence Interval for Starbucks Preference

  1. Standard Error (SE):

    • SE=extsqrtracp(1p)n=extsqrtrac0.51imes0.49100ext0.05SE = ext{sqrt} rac{p(1 - p)}{n} = ext{sqrt} rac{0.51 imes 0.49}{100} ext{≈} 0.05

  2. Margin of Error (ME):

    • At 95% confidence ($z = 1.96$):
      ME=1.96imes0.05ext0.10ME = 1.96 imes 0.05 ext{≈} 0.10 (or 10%)

  3. Confidence Interval Interpretation:

    • We are 95% confident that the true population preference for Starbucks lies between (51% - 10%) 41% and (51% + 10%) 61%.

Key Insights and Interpretation

Analysis Insights

  • The Lead: Starbucks holds a 12-point lead over Tim Horton’s in this sample.

  • Neutral Variable Impact: The 10% ”No Preference” group indicates high brand engagement in this population.

  • Statistical Power Consideration: With $n = 100$, the margin of error (±10%) is considerable. A critical note is that lower bound of Starbucks CI (41%) overlaps with Tim Horton’s result (39%), necessitating inferential testing (like Chi-Square) to affirm significance.

More Detail on the Chi-Square Approach

Data Overview for Chi-Square Test

  • Total Sample Size ($n$): 100

  • Preference Observations:

    • Starbucks: 51%

    • Tim Horton’s: 39%

    • No Preference: 10%

The Research Question

  • Is the observed preference for Starbucks statistically significant compared to a neutral distribution?

The Chi-Square Goodness-of-Fit Test

Null Hypothesis Formulation

  • Null Hypothesis ($H_0$): All preferences are equal.

  • Chi-Square Test Formula: extχ2=extΣrac(O<em>iE</em>i)2Eiext{χ}^2 = ext{Σ} rac{(O<em>i - E</em>i)^2}{E_i}

    • Where:

    • $O_i$: Observed frequency

    • $E_i$: Expected frequency (33.33 for each category)

Calculation Results of Chi-Square Test

Contributions by Category

Category

Observed

Expected

Contribution

Starbucks

51

33.33

9.37

Tim Horton’s

39

33.33

0.96

No Preference

10

33.33

16.33

Total

100

100

26.66

Degrees of Freedom Calculation

  • Degrees of Freedom ($df$): $3 - 1 = 2$

Conclusion of Chi-Square Test

Final Results and Implications

  • Calculated Chai-Square ($ ext{χ}^2$): 26.66

  • Critical Value ($ ext{α} = 0.05$): 5.99

  • Result: Since 26.66 > 5.99, we reject the Null Hypothesis. The distribution of coffee preferences is not random.

  • Note: The low ”No Preference” count indicates that consumers have strong opinions regarding these brands.

    • Note: Coffee was mentioned abstractly here; the initial context was about a sample preference inquiry.