3.1.1 Example 1: One-Sample Z-Test

A company claims its energy bars have an average of 20 grams of protein.The population standard deviation is known to be 1.5 grams. A random sample of 35 bars is taken, and the sample mean is found to be 20.6 grams. At a 5% significance level, is there evidence that the mean protein content is different from 20 grams?

Step 1: Hypotheses

• Null hypothesis:
  \(H_0: \mu = 20\)

• Alternative hypothesis:
  \(H_1: \mu \neq 20\) (two-tailed test)

Step 2: Significance Level \(\alpha = 0.05\)

Step 3: Test Statistic (Manual Calculation)

Given:

• Sample mean \(\bar{x} = 20.6\)
  
• Population mean \(\mu_0 = 20\)
  
• Population standard deviation \(\sigma = 1.5\)
  
• Sample size \(n = 35\)

\[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]

\[ z = \frac{20.6 - 20}{1.5 / \sqrt{35}} \]

\[ z = \frac{0.6}{1.5 / 5.916} = \frac{0.6}{0.2535} \approx 2.37 \]

Step 4: P-value

For a two-tailed test:

\[ p\text{-value} = 2 \times P(Z > |2.37|) \]

From Z-table: \(P(Z > 2.37) \approx 0.0089\)

\[ p\text{-value} = 2 \times 0.0089 = 0.0178 \]

Step 5: Decision

• \(p\)-value (0.0178) < \(\alpha\) (0.05)
  
• Reject \(H_0\)

Step 6: Conclusion

At the 5% significance level, there is sufficient evidence to conclude that the true mean protein content of the energy bars is different from 20 grams.

Verification in R:

# Given values
xbar <- 20.6
mu0 <- 20
sigma <- 1.5
n <- 35

# Z statistic
z <- (xbar - mu0) / (sigma / sqrt(n))
z

## [1] 2.366432

# Two-tailed p-value
p_value <- 2 * (1 - pnorm(abs(z)))
p_value

## [1] 0.01796048

3.1.2 Example 2: Two tailed test

A company claims its light bulbs last 1200 hours. The population standard deviation is 100 hours. A sample of 50 bulbs has a mean lifespan of 1175 hours. Test at \(\alpha=0.05\).

Step 1: Hypotheses
\(H_0: \mu = 1200\)
\(H_1: \mu \neq 1200\)

Step 2: Given Values
\(\mu_0=1200\), \(\sigma=100\), \(\bar{x}=1175\), \(n=50\), \(\alpha=0.05\)

Step 3: Test Statistic
\(SE = 100/\sqrt{50} = 14.14\)
\(Z = (1175-1200)/14.14 = -25/14.14 = -1.77\)

Step 4: Critical Values
Two-tailed test, \(Z_{0.025}=\pm1.96\)

Step 5: Decision
\(-1.77 > -1.96\) → Fail to reject \(H_0\).

Step 6: Conclusion
No significant evidence that mean lifespan differs from 1200 hours (\(Z=-1.77\), \(p>0.05\)).

3.1.3 Example 3: One-Tailed Test

A cereal company claims boxes contain at least 500 g. Population \(\sigma=15\). A sample of 40 boxes has mean weight 495 g. Test at \(\alpha=0.01\) if boxes are underfilled.

Step 1: Hypotheses
\(H_0: \mu \geq 500\)
\(H_1: \mu < 500\)

Step 2: Given Values
\(\mu_0=500\), \(\sigma=15\), \(\bar{x}=495\), \(n=40\), \(\alpha=0.01\)

Step 3: Test Statistic
\(SE = 15/\sqrt{40} = 2.37\)
\(Z = (495-500)/2.37 = -5/2.37 = -2.11\)

Step 4: Critical Value
Left-tailed test, \(Z_{0.01}=-2.33\)

Step 5: Decision
\(-2.11 > -2.33\) → Fail to reject \(H_0\).

Step 6: Conclusion
No significant evidence that boxes are underfilled (\(Z=-2.11\), \(p>0.01\)).

3.1.4 Example 4: Right-Tailed Test

A school district claims average SAT math score is 520 with \(\sigma=100\). A new teaching method is tested on 64 students, yielding mean score 540. Test at \(\alpha=0.05\) if the new method improves scores.

Step 1: Hypotheses
\(H_0: \mu \leq 520\)
\(H_1: \mu > 520\)

Step 2: Given Values
\(\mu_0=520\), \(\sigma=100\), \(\bar{x}=540\), \(n=64\), \(\alpha=0.05\)

Step 3: Test Statistic
\(SE = 100/\sqrt{64} = 12.5\)
\(Z = (540-520)/12.5 = 20/12.5 = 1.60\)

Step 4: Critical Value
Right-tailed test, \(Z_{0.05}=1.645\)

Step 5: Decision
\(1.60 < 1.645\) → Fail to reject \(H_0\).

Step 6: Conclusion
No significant evidence that the new method improves scores (\(Z=1.60\), \(p>0.05\)).

• When to use:
  • Z-test: Population standard deviation (\(\sigma\)) known
    
  • t-test: Population standard deviation unknown (use sample \(s\))

3.2.1 Example 1: One-Sample t-test (Practical with R & Python)

Problem: A car manufacturer claims a new model gets at least 40 MPG. A consumer agency tests a random sample of 12 cars, with the following results:

39.2, 40.5, 38.7, 41.0, 39.8, 40.9, 38.5, 39.9, 40.2, 39.3, 41.1, 38.6

Test the manufacturer’s claim at α = 0.05. Assume MPG is approximately normally distributed.

Solution (Manual Steps First):

1. Hypotheses:

H₀: μ ≥ 40 (Manufacturer's claim is true)

H₁: μ < 40 (The mean MPG is less than 40) -> One-tailed (left-tailed) test

2. Significance Level: α = 0.05

3. Calculate Sample Statistics:

Calculate the mean (x̄) and standard deviation (s) of the sample data

\[ \bar{x} = \frac{\text{sum of all values}}{12} \approx 39.783 \]

s ≈ 0.905 (calculated using the sample standard deviation formula)

5. Test Statistic:

\[ t = \frac{\bar{x} - \mu_{0}}{s / \sqrt{n}} = \frac{39.783 - 40}{0.905 / \sqrt{12}} = \frac{-0.217}{0.905 / 3.464} = \frac{-0.217}{0.261} \approx -0.831 \]

df = n - 1 = 11

Find p-value (using t-table for df=11):

This is a left-tailed test. We need P(T < -0.831).

From a t-table, for df=11, the value 0.831 falls between 0.697 and 1.363. The corresponding one-tailed probabilities are 0.25 and 0.10.

We can estimate p-value > 0.10. (Software will give a more precise value).

6. Decision (Manual):

Our estimated p-value (> 0.10) is greater than α (0.05). Therefore, we fail to reject H₀.

Now, let’s solve it precisely with code:

In R:

# Sample data
mpg_data <- c(39.2, 40.5, 38.7, 41.0, 39.8, 40.9, 38.5, 39.9, 40.2, 39.3, 41.1, 38.6)

# Perform one-sample t-test (alternative="less" for H1: mu < 40)
test_result <- t.test(mpg_data, mu = 40, alternative = "less")

# Print the results
print(test_result)

## 
##  One Sample t-test
## 
## data:  mpg_data
## t = -0.69814, df = 11, p-value = 0.2498
## alternative hypothesis: true mean is less than 40
## 95 percent confidence interval:
##      -Inf 40.30138
## sample estimates:
## mean of x 
##  39.80833

Conclusion from R:

The p-value is 0.2116. Since 0.2116 > 0.05, we fail to reject H₀. There is not enough evidence to reject the manufacturer’s claim that the mean MPG is at least 40.

In Python:

#import numpy as np
#from scipy import stats

# Sample data
#mpg_data = np.array([39.2, 40.5, 38.7, 41.0, 39.8, 40.9, 38.5, 39.9, 40.2, 39.3, 41.1, 38.6])

# Perform one-sample t-test
# 'alternative="less"' sets H1: mu < 40
#t_stat, p_value = stats.ttest_1samp(mpg_data, popmean=40, alternative='less')

# Print the results
#print(f"t-statistic: {t_stat:.4f}")
#print(f"p-value: {p_value:.4f}")
# For a one-tailed test, the p-value from `ttest_1samp` with 'alternative' is already correct.
# If using an older version without 'alternative', p_value / 2 for one-tailed.

Conclusion is the same as in R.

3.2.2 Example 2: Two-tailed Test (Fail to Reject)

Problem: A machine claims to fill jars with 100 g. A sample of \(n=10\) jars has \(\bar{x}=105\) g and \(s=8\) g. Test \(H_0: \mu=100\) vs \(H_1: \mu \neq 100\) at \(\alpha=0.05\).

Step 1: State hypotheses
\(H_0: \mu=100\)
\(H_1: \mu\neq100\)

Step 2: Compute standard error
\(SE = s/\sqrt{n} = 8/\sqrt{10} = 8/3.1623 = 2.5298\)

Step 3: Compute test statistic
\(t = (\bar{x}-\mu_0)/SE = (105-100)/2.5298 = 5/2.5298 = 1.976\)

Step 4: Degrees of freedom
\(df = 10-1=9\)

Step 5: Critical value
\(t_{0.975,9}=2.262\)

Decision: \(|t|=1.976 < 2.262\), so fail to reject \(H_0\). No evidence mean differs from 100.

In R:

n <- 10
xbar <- 105
mu0 <- 100
s <- 8

se <- s/sqrt(n)
t_stat <- (xbar - mu0)/se
df <- n-1
p_val <- 2*pt(-abs(t_stat), df)

list(t_statistic = t_stat, df = df, p_value = p_val)

## $t_statistic
## [1] 1.976424
## 
## $df
## [1] 9
## 
## $p_value
## [1] 0.07951604

3.2.3 Example 3: Two-tailed Test with 95% CI (Reject \(H_0\))

Problem: \(n=15\), \(\bar{x}=52\), \(s=3.5\), test \(H_0: \mu=50\) vs \(H_1: \mu\neq50\).

Step 1: SE
\(SE = 3.5/\sqrt{15} = 3.5/3.873 = 0.9035\)

Step 2: Test statistic
\(t = (52-50)/0.9035 = 2/0.9035 = 2.214\)

Step 3: df
\(df=14\)

Step 4: Critical value
\(t_{0.975,14}=2.145\)

Decision: \(t=2.214 > 2.145\), reject \(H_0\).

Step 5: 95% CI
Margin = \(t_{0.975,14}\times SE = 2.145\times 0.9035 = 1.939\)
CI = \(52\pm1.939 = (50.06,53.94)\)

Conclusion: Reject \(H_0\) (p < 0.05). The mean sodium content differs from 50 mg. The 95% CI provides the plausible range.

In R:

n <- 15
xbar <- 52
mu0 <- 50
s <- 3.5

se <- s/sqrt(n)
t_stat <- (xbar - mu0)/se
df <- n-1
p_val <- 2*pt(-abs(t_stat), df)

# 95% CI
alpha <- 0.05
t_crit <- qt(1-alpha/2, df)
margin <- t_crit*se
ci <- c(xbar - margin, xbar + margin)

list(t_statistic = t_stat, df = df, p_value = p_val, CI_95 = ci)

## $t_statistic
## [1] 2.213133
## 
## $df
## [1] 14
## 
## $p_value
## [1] 0.04400273
## 
## $CI_95
## [1] 50.06176 53.93824

3.2.4 Example 4: One-sided Test with 99% CI (Reject \(H_0\))

Problem: Training course improvement, \(n=25\), \(\bar{x}=4\), \(s=6\), test \(H_0: \mu=0\) vs \(H_1: \mu>0\) at \(\alpha=0.01\).

Step 1: SE
\(SE = 6/\sqrt{25} = 6/5 = 1.2\)

Step 2: Test statistic
\(t = (4-0)/1.2 = 4/1.2 = 3.333\)

Step 3: df
\(df=24\)

Step 4: Critical value (one-sided)
\(t_{0.99,24}=2.492\)

Decision: \(t=3.333 > 2.492\), reject \(H_0\). Strong evidence of positive improvement.

Step 5: 99% CI
Margin = \(t_{0.995,24}\times SE = 2.797\times1.2=3.356\)
CI = \(4\pm3.356 = (0.64,7.36)\)

Conclusion: Strong evidence (\(p<0.01\)) that the course increases scores. The 99% CI shows the true mean improvement is positive.

In R:

n <- 25
xbar <- 4
mu0 <- 0
s <- 6

se <- s/sqrt(n)
t_stat <- (xbar - mu0)/se
df <- n-1
p_val <- 1 - pt(t_stat, df) # one-sided

# 99% CI
t_crit <- qt(0.995, df)
margin <- t_crit*se
ci <- c(xbar - margin, xbar + margin)

list(t_statistic = t_stat, df = df, p_value = p_val, CI_99 = ci)

## $t_statistic
## [1] 3.333333
## 
## $df
## [1] 24
## 
## $p_value
## [1] 0.001388157
## 
## $CI_99
## [1] 0.6436726 7.3563274

5.1.1 Example 1: Two-Tailed Z-Test

A study compares test scores between two schools.

• School A: \(n_1 = 40, \ \bar{x}_1 = 85, \ \sigma_1 = 8\)
  
• School B: \(n_2 = 50, \ \bar{x}_2 = 82, \ \sigma_2 = 7.5\)
  
• Significance level: \(\alpha = 0.05\)

Solution

Step 1: State Hypotheses

\[H_0: \mu_1 = \mu_2 \quad \text{(no difference in means)}\]

\[H_1: \mu_1 \neq \mu_2 \quad \text{(means differ)}\]

Step 2: Compute Test Statistic (Manual Calculation)

The standard error (SE) is:

\[SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{8^2}{40} + \frac{7.5^2}{50}}= \sqrt{\frac{64}{40} + \frac{56.25}{50}}= \sqrt{1.6 + 1.125} = \sqrt{2.725} \approx 1.651\]

The z-statistic is:

\[z = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{85 - 82}{1.651} \approx 1.82\]

Step 3: Critical Value & Decision Rule

For a two-tailed test at \(\alpha = 0.05\):

\[z_{\alpha/2} = \pm 1.96\]

Decision rule:
- If \(|z| > 1.96\), reject \(H_0\).
- Otherwise, fail to reject \(H_0\).

Here, \(|1.82| < 1.96\), so we fail to reject \(H_0\).

Step 4: Conclusion

There is no significant difference between the mean test scores of the two schools.

• Test statistic: \(z = 1.82\)
  
• Critical values: \(\pm 1.96\)
  
• Decision: Fail to reject \(H_0\)
  
• Interpretation: The evidence is insufficient at \(\alpha = 0.05\) to conclude a difference in mean test scores.

Step 5: R Verification

# Given data
x1 <- 85; x2 <- 82
n1 <- 40; n2 <- 50
sigma1 <- 8; sigma2 <- 7.5

# Standard error
SE <- sqrt((sigma1^2/n1) + (sigma2^2/n2))
SE

## [1] 1.650757

# Z statistic
z_value <- (x1 - x2)/SE
z_value

## [1] 1.817348

# Two-tailed p-value
p_value <- 2 * (1 - pnorm(abs(z_value)))
p_value

## [1] 0.06916391

5.1.2 Example 2: One-Tailed Test (Right-Tailed)

A company tests two production methods. Method X (n=35) has mean output=120 units/hour, σ=10. Method Y (n=40) has mean output=115 units/hour, σ=9. Test at α=0.01 if Method X is superior.

Solution

Step 1: State Hypotheses

• Null Hypothesis (H₀): μₓ ≤ μᵧ (Method X is not superior)

• Alternative Hypothesis (H₁): μₓ > μᵧ (Method X is superior)

Step 2: Given Values - Method X:
- Sample size (n₁) = 35
- Mean (x̄₁) = 120
- Standard deviation (σ₁) = 10

• Method Y:
  • Sample size (n₂) = 40
    
  • Mean (x̄₂) = 115
    
  • Standard deviation (σ₂) = 9
    
• Significance level (α) = 0.01

Step 3: Calculate Z-Statistic

\[Z = \frac{x̄₁ - x̄₂}{\sqrt{\frac{σ₁^2}{n₁} + \frac{σ₂^2}{n₂}}} = \frac{120 - 115}{\sqrt{\frac{100}{35} + \frac{81}{40}}} = \frac{5}{\sqrt{2.857 + 2.025}} = \frac{5}{\sqrt{4.882}} = \frac{5}{2.210} ≈ 2.26 \]

Step 4: Critical Value

• Right-tailed test at α = 0.01
  
• Critical Z-value: Zₐ = 2.33

Step 5: Compare and Decide

• Since 2.26 < 2.33, the Z-statistic is not in the rejection region
  
• Fail to reject H₀

Step 6: Conclusion

There is no significant evidence that Method X is superior.
Z = 2.26, p > 0.01

5.1.3 Example 3: One-Tailed Test (Left-Tailed)

A nutritionist compares calorie intake between two diets. Diet A (n=60) has mean=1800 calories, σ=150. Diet B (n=55) has mean=1850 calories, σ=140. Test at α=0.05 if Diet A has lower calorie intake.

Solution

Step 1: State Hypotheses

• Null Hypothesis (H₀): μₐ ≥ μᵦ (Diet A is not lower in calories)

• Alternative Hypothesis (H₁): μₐ < μᵦ (Diet A has lower calorie intake)

Step 2: Given Values

• Diet A:
  • Sample size (n₁) = 60
    
  • Mean (x̄₁) = 1800
    
  • Standard deviation (σ₁) = 150
• Diet B:
  • Sample size (n₂) = 55
    
  • Mean (x̄₂) = 1850
    
  • Standard deviation (σ₂) = 140
    
• Significance level (α) = 0.05

Step 3: Calculate Z-Statistic

\[ Z = \frac{x̄₁ - x̄₂}{\sqrt{\frac{σ₁^2}{n₁} + \frac{σ₂^2}{n₂}}} = \frac{1800 - 1850}{\sqrt{\frac{22500}{60} + \frac{19600}{55}}} = \frac{-50}{\sqrt{375 + 356.36}} = \frac{-50}{\sqrt{731.36}} = \frac{-50}{27.04} ≈ -1.85 \]

Step 4: Critical Value

• Left-tailed test at α = 0.05
  
• Critical Z-value: Zₐ = -1.645

Step 5: Compare and Decide

• Since -1.85 < -1.645, the Z-statistic is in the rejection region
  
• Reject H₀

Step 6: Conclusion

There is significant evidence that Diet A has lower calorie intake.
Z = -1.85, p < 0.05

5.2.1 Purpose

A two-tailed test is used to determine whether there is a significant difference between the means of two populations. Unlike a one-tailed test, which focuses on whether one mean is greater than or less than the other, a two-tailed test checks for any difference (positive or negative).

5.2.2 Types of Two-Sample Tests

• Independent Samples: The two samples are unrelated.

• Paired Samples: The two samples are related (e.g., before-and-after measurements on the same subjects).

5.2.3 Assumptions

• The samples are randomly selected.

• Independence of observations

• Equal variances for pooled t-test

• Random sampling

• The populations are normally distributed (or sample sizes are large enough for the Central Limit Theorem to apply).

• For independent samples: The variances of the two populations may or may not be equal.

• For paired samples: The differences between paired observations are normally distributed.

5.2.4 Hypotheses

For independent samples:

• Null hypothesis (\(H_0\)): \(\mu_1 = \mu_2\) (no difference in population means)

• Alternative hypothesis (\(H_a\)): \(\mu_1 \neq \mu_2\) (there is a difference)

For paired samples:

• Null hypothesis (\(H_0\)): \(\mu_d = 0\) (mean difference is zero)

• Alternative hypothesis (\(H_a\)): \(\mu_d \neq 0\) (mean difference is not zero)

5.2.5 Test Statistics

For Independent Samples

• If population variances are assumed equal: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{S_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}} \]

  where \(S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}\) is the pooled variance.

• If population variances are not assumed equal (Welch’s t-test): \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} \] Degrees of freedom are calculated using the Welch-Satterthwaite equation.

For Paired Samples \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \] where \(\bar{d}\) is the mean of the differences, \(s_d\) is the standard deviation of the differences, and \(n\) is the number of pairs.

5.2.6 Decision Rule

• Compare the calculated \(t\)-value with the critical \(t\)-value from the \(t\)-distribution table at the given significance level (\(\alpha\)) and degrees of freedom.

• Reject \(H_0\) if the absolute value of the calculated \(t\)-value exceeds the critical \(t\)-value.

There are two types: independent (unpaired) and paired t-tests.

Two-Sample Pooled t-Test (Equal Variances): Used to compare the means of two independent samples when population variances are assumed equal.

Test Statistic

\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]

Pooled Variance

\[s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\]

Degrees of Freedom

\[df = n_1 + n_2 - 2\]

Welch’s t-Test (Unequal Variances): Used to compare the means of two independent samples when population variances are not assumed equal.

Test Statistic

\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

Degrees of Freedom (Welch-Satterthwaite Approximation)

\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]

Paired t-Test: Used to compare means from the same group at different times or under different conditions.

Test Statistic

\[t = \frac{\bar{d}}{s_d / \sqrt{n}}\]

Where:

• \(\bar{d}\) = mean of the differences
  
• \(s_d\) = standard deviation of the differences

Degrees of Freedom

\[df = n - 1\] (n = No. of paires)

I will provide examples of both.

• Example 1: Independent two-sample t-test (equal variances assumed)

• Example 2: Independent two-sample t-test (Welch’s t-test, unequal variances not assumed)

• Example 3: Paired two-sample t-test

5.2.7 Solved Example 1: Independent Samples**

Problem: A researcher wants to compare the average test scores of students from two different teaching methods. A random sample of 25 students from Method A has a mean score of 78 with a standard deviation of 10. A random sample of 30 students from Method B has a mean score of 75 with a standard deviation of 12. Assume unequal variances. Perform a two-tailed test at \(\alpha = 0.05\).

Solution:

1. State the hypotheses:

   • \(H_0: \mu_A = \mu_B\)
   • \(H_a: \mu_A \neq \mu_B\)

2. Calculate the test statistic:

   Using Welch’s \(t\)-test: \[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{S_A^2}{n_A} + \frac{S_B^2}{n_B}}} \] Substituting values: \[ t = \frac{78 - 75}{\sqrt{\frac{10^2}{25} + \frac{12^2}{30}}} = \frac{3}{\sqrt{4 + 4.8}} = \frac{3}{\sqrt{8.8}} \approx \frac{3}{2.97} \approx 1.01 \]

3. Degrees of freedom:

   Using the Welch-Satterthwaite formula: \[ df \approx \frac{\left(\frac{S_A^2}{n_A} + \frac{S_B^2}{n_B}\right)^2}{\frac{\left(\frac{S_A^2}{n_A}\right)^2}{n_A - 1} + \frac{\left(\frac{S_B^2}{n_B}\right)^2}{n_B - 1}} \] Substituting values: \[ df \approx \frac{(4 + 4.8)^2}{\frac{4^2}{24} + \frac{4.8^2}{29}} \approx \frac{8.8^2}{\frac{16}{24} + \frac{23.04}{29}} \approx \frac{77.44}{0.667 + 0.795} \approx 52 \]

4. Critical value:

   From the \(t\)-table, \(t_{\text{critical}} = 2.009\) (for \(\alpha = 0.05\) and \(df = 52\)).

5. Decision:

   Since \(|t| = 1.01 < 2.009\), we fail to reject \(H_0\). There is no significant difference in the means.

5.2.8 Solved Example 2: Paired Samples

Problem: A study measures the blood pressure of 10 patients before and after a new medication. The differences in systolic blood pressure are: \([-5, -3, -8, -6, -4, -7, -2, -5, -6, -4]\). Perform a two-tailed test at \(\alpha = 0.05\).

Solution: 1. State the hypotheses: - \(H_0: \mu_d = 0\) - \(H_a: \mu_d \neq 0\)

2. Calculate the mean and standard deviation of differences:

   • Mean: \(\bar{d} = \frac{-5 - 3 - 8 - 6 - 4 - 7 - 2 - 5 - 6 - 4}{10} = -5\)
   • Standard deviation: \(s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{(-5+5)^2 + (-3+5)^2 + ... + (-4+5)^2}{9}} = \sqrt{\frac{50}{9}} \approx 2.36\)

3. Calculate the test statistic: \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{-5}{2.36 / \sqrt{10}} = \frac{-5}{2.36 / 3.16} = \frac{-5}{0.747} \approx -6.7 \]

4. Degrees of freedom: \(df = n - 1 = 10 - 1 = 9\)

5. Critical value: From the \(t\)-table, \(t_{\text{critical}} = 2.262\) (for \(\alpha = 0.05\) and \(df = 9\)).

6. Decision: Since \(|t| = 6.7 > 2.262\), we reject \(H_0\). There is a significant difference in blood pressure before and after the medication.

5.2.9 Solved Example 3: Independent Samples (Equal Variances)

Problem: A researcher wants to compare the average heights of plants grown in two different fertilizers. A random sample of 15 plants from Fertilizer A has a mean height of 20 cm with a standard deviation of 3 cm. A random sample of 18 plants from Fertilizer B has a mean height of 18 cm with a standard deviation of 4 cm. Assume equal variances. Perform a two-tailed test at \(\alpha = 0.05\).

Solution:

1. State the hypotheses:

   • \(H_0: \mu_A = \mu_B\)
   • \(H_a: \mu_A \neq \mu_B\)

2. Calculate the pooled variance: \[ S_p^2 = \frac{(n_A - 1)S_A^2 + (n_B - 1)S_B^2}{n_A + n_B - 2} \] Substituting values: \[ S_p^2 = \frac{(15 - 1)(3^2) + (18 - 1)(4^2)}{15 + 18 - 2} = \frac{14(9) + 17(16)}{31} = \frac{126 + 272}{31} = \frac{398}{31} \approx 12.84 \]

3. Calculate the test statistic: \[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{S_p^2 \left( \frac{1}{n_A} + \frac{1}{n_B} \right)}} \] Substituting values: \[ t = \frac{20 - 18}{\sqrt{12.84 \left( \frac{1}{15} + \frac{1}{18} \right)}} = \frac{2}{\sqrt{12.84 \left( 0.0667 + 0.0556 \right)}} = \frac{2}{\sqrt{12.84 \cdot 0.1223}} = \frac{2}{\sqrt{1.57}} \approx \frac{2}{1.25} \approx 1.6 \]

4. Degrees of freedom: \[ df = n_A + n_B - 2 = 15 + 18 - 2 = 31 \]

5. Critical value: From the \(t\)-table, \(t_{\text{critical}} = 2.042\) (for \(\alpha = 0.05\) and \(df = 31\)).

6. Decision: Since \(|t| = 1.6 < 2.042\), we fail to reject \(H_0\). There is no significant difference in the mean heights of plants grown with the two fertilizers.

5.2.10 Solved Example 4: Paired Samples

Problem: A study measures the reaction times of 12 drivers before and after consuming alcohol. The differences in reaction times (in seconds) are: \([0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3]\). Test if alcohol significantly increases reaction time at \(\alpha = 0.01\).

Solution:

1. State the hypotheses:

   • \(H_0: \mu_d = 0\)
   • \(H_a: \mu_d \neq 0\)

2. Calculate the mean and standard deviation of differences:

   • Mean: \(\bar{d} = \frac{0.2 + 0.3 + 0.4 + ... + 1.3}{12} = \frac{7.8}{12} = 0.65\)

   • Standard deviation: \(s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}\) \[ s_d = \sqrt{\frac{(0.2 - 0.65)^2 + (0.3 - 0.65)^2 + ... + (1.3 - 0.65)^2}{11}} \]

     \[ = \sqrt{\frac{0.2025 + 0.1225 + ... + 0.4225}{11}} \] = $$

3. Calculate the test statistic: \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{0.65}{0.51 / \sqrt{12}} = \frac{0.65}{0.51 / 3.46} = \frac{0.65}{0.147} \approx 4.42 \]

4. Degrees of freedom: \[ df = n - 1 = 12 - 1 = 11 \]

5. Critical value: From the \(t\)-table, \(t_{\text{critical}} = 3.106\) (for \(\alpha = 0.01\) and \(df = 11\)).

6. Decision: Since \(|t| = 4.42 > 3.106\), we reject \(H_0\). Alcohol significantly increases reaction time.

5.2.11 Solved Example 5: Independent Samples (Unequal Variances)

Problem: A company compares the productivity of two teams. Team X produces 30 items with a mean output of 50 units and a standard deviation of 8 units. Team Y produces 25 items with a mean output of 45 units and a standard deviation of 10 units. Assume unequal variances. Perform a two-tailed test at \(\alpha = 0.05\).

Solution:

1. State the hypotheses:

   • \(H_0: \mu_X = \mu_Y\)
   • \(H_a: \mu_X \neq \mu_Y\)

2. Calculate the test statistic: Using Welch’s \(t\)-test: \[ t = \frac{\bar{x}_X - \bar{x}_Y}{\sqrt{\frac{S_X^2}{n_X} + \frac{S_Y^2}{n_Y}}} \] Substituting values: \[ t = \frac{50 - 45}{\sqrt{\frac{8^2}{30} + \frac{10^2}{25}}} \] \[ = \frac{5}{\sqrt{\frac{64}{30} + \frac{100}{25}}} = \frac{5}{\sqrt{2.13 + 4}} \]

   \[ = \frac{5}{\sqrt{6.13}} \approx \frac{5}{2.47} \approx 2.02 \]

3. Degrees of freedom: Using the Welch-Satterthwaite formula: \[ df \approx \frac{\left(\frac{S_X^2}{n_X} + \frac{S_Y^2}{n_Y}\right)^2}{\frac{\left(\frac{S_X^2}{n_X}\right)^2}{n_X - 1} + \frac{\left(\frac{S_Y^2}{n_Y}\right)^2}{n_Y - 1}} \] Substituting values: \[ df \approx \frac{(2.13 + 4)^2}{\frac{2.13^2}{29} + \frac{4^2}{24}} \]

   \[ = \frac{6.13^2}{\frac{4.54}{29} + \frac{16}{24}}\] \[ = \frac{37.57}{0.157 + 0.667} \approx \frac{37.57}{0.824} \approx 45.6 \]

4. Critical value: From the \(t\)-table, \(t_{\text{critical}} = 2.014\) (for \(\alpha = 0.05\) and \(df = 45\)).

5. Decision: Since \(|t| = 2.02 > 2.014\), we reject \(H_0\). There is a significant difference in productivity between the two teams.

5.2.12 Solved Example 6: Paired Samples

Problem: A study measures the cholesterol levels of 8 patients before and after a new diet. The differences in cholesterol levels are: \([-10, -15, -20, -12, -18, -14, -16, -13]\). Test if the diet significantly reduces cholesterol at \(\alpha = 0.05\).

Solution:

1. State the hypotheses:

   • \(H_0: \mu_d = 0\)
   • \(H_a: \mu_d \neq 0\)

2. Calculate the mean and standard deviation of differences:

   • Mean: \(\bar{d} = \frac{-10 - 15 - 20 - 12 - 18 - 14 - 16 - 13}{8} = \frac{-118}{8} = -14.75\)
   • Standard deviation: \(s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}\) \[ s_d = \sqrt{\frac{(-10 + 14.75)^2 + (-15 + 14.75)^2 + ... + (-13 + 14.75)^2}{7}} \] \[ = \sqrt{\frac{22.56 + 0.06 + ... + 3.06}{7}} = \sqrt{\frac{33.63}{7}} \approx \sqrt{4.8} \approx 2.19 \]

3. Calculate the test statistic: \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{-14.75}{2.19 / \sqrt{8}} \] \[ = \frac{-14.75}{2.19 / 2.83} = \frac{-14.75}{0.774} \approx -19.05 \]

4. Degrees of freedom: \[ df = n - 1 = 8 - 1 = 7 \]

5. Critical value: From the \(t\)-table, \(t_{\text{critical}} = 2.365\) (for \(\alpha = 0.05\) and \(df = 7\)).

6. Decision: Since \(|t| = 19.05 > 2.365\), we reject \(H_0\). The diet significantly reduces cholesterol levels.

5.2.13 Solved Example 7: Two-Tailed Test (Equal Variances Assumed)

A researcher wants to compare the effectiveness of two teaching methods. Method A is used on 25 students, Method B on 30 students. Test scores are recorded:

Method A: 78, 82, 85, 79, 83, 88, 76, 81, 84, 80, 82, 85, 79, 83, 87, 77, 82, 86, 80, 84, 78, 81, 85, 79, 83

Method B: 75, 79, 82, 76, 80, 84, 74, 78, 81, 77, 79, 83, 75, 80, 82, 76, 79, 81, 77, 80, 75, 78, 82, 76, 79, 81, 75, 78, 82, 76

Test at α = 0.05 if there’s a significant difference between methods.

Solution

Step 1: State Hypotheses

• H₀: μ₁ = μ₂ (No difference in mean scores)

• H₁: μ₁ ≠ μ₂ (Means differ significantly)

Step 2: Sample Statistics

• Method A: n₁ = 25, x̄₁ = 81.6, s₁ = 3.24
  
• Method B: n₂ = 30, x̄₂ = 78.9, s₂ = 2.98

Step 3: Pooled Variance

\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{(24)(10.50) + (29)(8.88)}{53} = \frac{252 + 257.5}{53} = 9.61 \]

\[ s_p = \sqrt{9.61} = 3.10 \]

Step 4: t-Statistic

\[ t = \frac{x̄_1 - x̄_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{2.7}{3.10 \times \sqrt{0.04 + 0.0333}} = \frac{2.7}{0.84} ≈ 3.21 \]

Step 5: Critical Value

• Degrees of freedom: df = 25 + 30 - 2 = 53

• α = 0.05 (two-tailed)

• Critical t-value: ±2.006

Step 6: Decision

• Since 3.21 > 2.006, reject H₀

Step 7: Conclusion

There is a significant difference in effectiveness between the two teaching methods.
t(53) = 3.21, p < 0.05

5.2.14 Solved Example 8: One-Tailed Test (Welch’s t-test, Unequal Variances)

A company tests two battery types. Type X (n=15) has mean life=120 hours, s=12. Type Y (n=20) has mean life=115 hours, s=8. Test at α=0.05 if Type X lasts longer.

Solution

Step 1: State Hypotheses - H₀: $ μ_X ≤ μ_Y $

• H₁: $ μ_X > μ_Y $

Step 2: Given Values - Type X: $n₁ = 15, x̄₁ = 120, s₁ = 12 $ - Type Y: \(n₂ = 20, x̄₂ = 115, s₂ = 8\)

Step 3: Welch’s t-Statistic

\[ t = \frac{x̄_1 - x̄_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{5}{\sqrt{9.6 + 3.2}} = \frac{5}{\sqrt{12.8}} = \frac{5}{3.58} ≈ 1.40 \]

Step 4: Degrees of Freedom

\[ df = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}} = \frac{(12.8)^2}{\frac{(9.6)^2}{14} + \frac{(3.2)^2}{19}} = \frac{163.84}{6.58 + 0.54} ≈ 23.01 \]

Step 5: Critical Value - df ≈ 23, α = 0.05 (one-tailed)
- Critical t-value: 1.714

Step 6: Decision - Since 1.40 < 1.714, fail to reject H₀

Step 7: Conclusion No significant evidence that Type X batteries last longer.
t(23) = 1.40, p > 0.05

5.2.15 Solved Example 9: Practical Application with Raw Data

A fitness trainer compares weight loss between two diets. Participants are randomly assigned:

Diet Plan A (n=12): 5.2, 4.8, 6.1, 5.5, 4.9, 5.8, 6.2, 5.1, 4.7, 5.9, 5.3, 5.6 kg

Diet Plan B (n=10): 4.1, 4.5, 3.9, 4.8, 4.0, 4.3, 3.7, 4.6, 4.2, 4.4 kg

Test at α=0.05 if Diet A leads to greater weight loss.

Solution

Step 1: State Hypotheses - H₀: μ_A ≤ μ_B - H₁: μ_A > μ_B

Step 2: Sample Statistics

• Diet A: n₁ = 12, ∑x = 65.1, x̄ = 5.425, ∑x² = 356.7 9

• Diet B: n₂ = 10, ∑x = 42.5, x̄ = 4.25, ∑x² = 182.0 9

Variance Calculations

\[ s_1^2 = \frac{∑x_1^2 - (∑x_1)^2 / n_1}{n_1 - 1} = \frac{356.79 - (65.1)^2 / 12}{11} = \frac{356.79 - 353.17}{11} = 0.329 \]

\[ s_2^2 = \frac{∑x_2^2 - (∑x_2)^2 / n_2}{n_2 - 1} = \frac{182.09 - (42.5)^2 / 10}{9} = \frac{182.09 - 180.63}{9} = 0.162 \]

Step 3: Pooled Variance

\[ s_p^2 = \frac{(11)(0.329) + (9)(0.162)}{20} = \frac{3.619 + 1.458}{20} = 0.254 \quad s_p = \sqrt{0.254} = 0.504 \]

Step 4: t-Statistic

\[ t = \frac{5.425 - 4.25}{0.504 \sqrt{\frac{1}{12} + \frac{1}{10}}} = \frac{1.175}{0.504 \times \sqrt{0.1833}} = \frac{1.175}{0.216} ≈ 5.44 \]

Step 5: Critical Value

• df = 12 + 10 - 2 = 20

• α = 0.05 (one-tailed)

• Critical t-value: 1.725

Step 6: Decision

• Since 5.44 > 1.725, reject H₀

Step 7: Conclusion

Diet A leads to significantly greater weight loss than Diet B.
t(20) = 5.44, p < 0.001

6.1.1 Question 1: Product Quality Comparison

A consumer protection agency wants to compare the average weight of cereal boxes from two different brands. The population standard deviations are known from historical data.

Known Information:

Brand A: Population σ = 15 grams

Brand B: Population σ = 12 grams

Sample sizes: n₁ = 50 boxes of Brand A, n₂ = 45 boxes of Brand B

Sample means: x̄₁ = 498 grams, x̄₂ = 505 grams

Required

• Test at α = 0.05 if there is a significant difference in average weights between the two brands.

• Calculate the 95% confidence interval for the difference in means.

• What sample size would be needed to detect a difference of 5 grams with 90% power?

6.1.2 Question 2: Manufacturing Process Evaluation

A factory has two production lines for making electrical components. The specification requires components to have a resistance of 100 ohms. Historical data shows known population standard deviations.

Data:

Line 1: n = 60, x̄ = 101.5 ohms, σ = 2.5 ohms

Line 2: n = 55, x̄ = 99.8 ohms, σ = 3.0 ohms

Required:

• Test at α = 0.01 if Line 1 produces components with significantly higher resistance than Line 2.

• Calculate the p-value for the test.

• What is the probability of Type II error if the true difference in means is 1 ohm?

6.2.1 Question 3: Teaching Method Effectiveness

A school district wants to compare the effectiveness of traditional teaching methods versus technology-enhanced methods. Students were randomly assigned to two groups.

Test Scores Data: Traditional Group (n = 25): 78, 82, 75, 85, 80, 79, 83, 76, 81, 84, 77, 82, 79, 83, 78, 81, 80, 84, 76, 82, 79, 83, 77, 81, 80

Technology Group (n = 28): 85, 88, 82, 90, 86, 84, 87, 83, 89, 85, 81, 88, 84, 86, 83, 87, 85, 89, 84, 87, 82, 88, 86, 85, 87, 84, 88, 86

Required:

• Perform a two-sample t-test at α = 0.05 to determine if technology-enhanced methods lead to higher scores.

• Check the assumption of equal variances using an F-test.

• Calculate Cohen’s d to measure effect size.

• Interpret the results in educational context.

6.2.2 Question 4: Drug Efficacy Study

A pharmaceutical company is testing a new drug for cholesterol reduction. Patients are randomly assigned to treatment and control groups.

Cholesterol Reduction (mg/dL): Treatment Group (n = 20): 25, 28, 32, 35, 29, 31, 27, 34, 30, 33, 26, 29, 32, 36, 28, 31, 34, 27, 30, 33

Control Group (n = 18): 18, 20, 22, 19, 21, 23, 17, 20, 24, 19, 22, 18, 21, 23, 20, 22, 19, 21

Required:

• Test at α = 0.01 if the treatment group shows significantly greater cholesterol reduction.

• Should you use pooled or Welch’s t-test? Justify your choice.

• Calculate the 99% confidence interval for the difference in means.

• What are the practical implications of your findings?

6.2.3 Question 5: Independent Samples

A company tests two production methods. Method X produces 100 items with a mean weight of 15 kg and a standard deviation of 2 kg. Method Y produces 120 items with a mean weight of 14.5 kg and a standard deviation of 2.5 kg. Test if the mean weights differ at \(\alpha = 0.01\).

6.2.4 Question 6: Paired Samples

A fitness program measures the weights of 15 participants before and after a 6-month training period. The differences in weights are: \([-3, -2, -4, -1, -5, -3, -2, -4, -3, -2, -1, -3, -4, -2, -3]\). Test if the program significantly reduces weight at \(\alpha = 0.05\).

6.2.5 Question 7: Paired Samples

A psychologist studies the effect of a new therapy on stress levels. The stress scores (on a scale of 0 to 100) of 12 patients before and after the therapy are as follows:

Perform a two-tailed test at \(\alpha = 0.01\) to determine if the therapy significantly reduces stress levels.

6.2.6 Question 8: Paired Samples

A nutritionist evaluates the effectiveness of a new diet plan on blood sugar levels. The blood sugar levels (in mg/dL) of 15 patients before and after the diet are recorded as follows:

Perform a two-tailed test at \(\alpha = 0.05\) to determine if the diet significantly lowers blood sugar levels.

6.2.7 Question 9: Challenge Problem

Compare the performance of two algorithms on 50 datasets. Algorithm A has a mean accuracy of 85% with a standard deviation of 5%, while Algorithm B has a mean accuracy of 87% with a standard deviation of 6%. Assume unequal variances. Perform a two-tailed test at \(\alpha = 0.05\).