# t-Test

** Published:**

This post covers t-Tests.

# t-Distribution

- Z - test works when we know $\mu$ and $\sigma$
- Use Samples
- How different a sample mean is from a population
- How different two sample means are from each other
- Two samples can be
- Independent
- Dependent

- Two samples can be

- Estimate Population Standard Deviation using sample standard deviation with Bessel’s correction
- Bessel’s correction is the use of $n − 1$ instead of $n$ in the formula for the sample variance and sample standard deviation, where $n$ is the number of observations in a sample.
- This method corrects the bias in the estimation of the population variance.
- It also partially corrects the bias in the estimation of the population standard deviation.
- However, the correction often increases the mean squared error in these estimations.
- This technique is named after Friedrich Bessel.

- To find out how typical or atypical (unusual) a sample mean - find its location on the distribution of sample means i.e. sampling distribution
- we can determine when we know population parameters, $\mu, \sigma$
- $ std~errro= \frac{\sigma}{\sqrt{n}}$
- $z = \frac{sample~mean - \mu}{std~error} = \frac{mean~difference}{std~error}$
- Std for Samples = $S = \sqrt{\frac{\Sigma(X_i - \bar{X})^2}{n-1}}$

- Standard Error depends on sample, we cannot use $\sigma$ if we have sample
- Thus, we have a new distribution that is more prone to error - t-Distribution
- more spread out and thicker in the tails than a normal distribution
- Since large sample sizes gives skinnier sampling distribution

- more spread out and thicker in the tails than a normal distribution
- What happens as n increases?
- The t-Distribution approaches to Normal Distribution
- The t-Distribution gets Skinnier tails
- $S \rightarrow \sigma$

# Degree of Freedom - Sample Standard Deviation

- We can pick a sample of size $n$ from population using $n$ degrees of freedom
- Now to compute Standard Deviation, we need sample mean
- $\bar{X} = \frac{X_1+X_2+X_3+…+X_n}{n}$
- $ X_1+X_2+X_3+…+X_n = n . \bar{X} $
- $n-1$ Degrees of Freedom
- We may vary $n-1$ values to keep the sum of these values as $n\bar{X}$
- $n-1$ is the effective sample size since only $n-1$ values are independent if we know the mean.
- $S = \sqrt{\frac{\Sigma(X_i - \bar{X})^2}{n-1}}$

- As degrees of freedom increases, the t-distribution better approxiamate the normal distribution

# t-Table

- https://naneja.github.io/python/t-table
- Questions
What’s the t-critical value for a one-tailed alpha level of 0.05 with 12 degrees of freedom.

- df = 11 and p = 0.05
- Ans t = 1.782

- df = 11 and p = 0.05
What are t-critical values for 2-tailed test with $\alpha = 0.05$ and sample size 30

- $df = 29,~ p = \pm 0.025$
- Ans: $\pm 2.045$

What are the limits for the right area of t-statistic when the sample size is 24 and the t-statistic is 2.45

- $df=23, t=2.45$

- $p = .02 ~\text{or}~ .01$

# t-Statistic

$t = \frac{\bar{X}-\mu_0}{\frac{S}{\sqrt{n}}}$

- The larger/smaller the value of $\bar{X}$, the stronger the evidence that $\mu > \mu_0$
- The larger/smaller the value of $\bar{X}$, the stronger the evidence that $\mu < \mu_0$
- The further the value of $\bar{X}$ from $\mu_0$ in either direction, the stronger/weaker the evidence that $\mu \ne \mu_0$

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# One Sample t-Test

$t = \frac{\bar{X}-\mu_0}{\frac{S}{\sqrt{n}}}$

\[H_0: \mu = \mu_0 \\\begin{align*} H_A &: \mu < \mu_0 \\ &: \mu > \mu_0 \\ &: \mu \ne \mu_0 \end{align*}\]$\alpha$ Levels (column levels of t-table)

- What will increase the t-Statistic
- Large difference between $\bar{X}$ and $\mu_0$
- Larger $n$
- Larger $S$
- Large Standard Error

- Larger t-Statistic
- => Lower probability of obtaining t-Statistic
- => Larger $\bar{X} - \mu_0$

# P-Value

Compute t-statistic

- $t = \frac{\bar{X}-\mu_0}{\frac{S}{\sqrt{n}}}$

One-tailed Test

- p-value is the probability
- above the t-Statistic if it’s positive, or
- below the t-Statistic if it’s negative

- p-value is the probability
Two-tailed Test

- p-value is the probability of the sum of both
- above the t-Statistic and
- below the t-Statistic

- p-value is the probability of the sum of both
Reject the Null when the p-value is less than the $\alpha$ level

# Example - Finches Beek Width

- Average known Beak Width = 6.07 mm
- $H_0: \mu = 6.07$
- $H_A: \mu \ne 6.07$
- Sample Size = 500
- Degrees of Freedom = 499

- Compute the sample mean and std dev from the sample dataset
- $\bar{X} = 6.470$
- $S = \sqrt{\frac{\Sigma(X_i - \bar{X})^2}{n-1}} = 0.396$

- t-Statistic
- $ t = \frac{6.47 - 6.07}{0.396/\sqrt{500}} = \frac{0.4}{0.0179} = 22.346$

- Table
- $df = 499,~ t = 22.346$, https://naneja.github.io/python/t-table
- $p = 0$
- Reject null
- probability of getting this t-value is very very small
- probability of getting the sample with beek width 6.47 from the population with mean 6.07 is very very small

# Example

- Sample = [5, 19, 11, 23, 12, 7, 3, 21]
Is this sample mean significantly different from 10 at an alpha level of 0.05?

- Different => two-tailed t-test
- $n=8,~ \bar{X} = 12.625, S = 7.6$
- $t = \frac{\bar{X} - 10}{\frac{S}{\sqrt{n}}} = \frac{12.625 - 10}{\frac{7.6}{\sqrt{8}}} = 0.977$
- $df=7, t=0.977$
- https://naneja.github.io/python/t-table
- Two Tail Test
- $ p = 0.18 + 0.18 = 0.36 > 0.05 $

- Fail to Reject
- Thus, $H_0: \mu = 10$

# Example

Mean Rent = 1830 for all apartments

Company A wants to know if the rent they are charging is significantly different at $\alpha = 0.05$

- Sample:
- $n=25,~ \bar{X}=1700,~ S=200$

- Sample:
Hypothesis

- $H_0: \mu = 1830$ and
- $H_A: \mu \ne 1830$
- What are t-critical values
- $t = \frac{\bar{X} - 1830}{\frac{S}{\sqrt{n}}} = \frac{1700 - 1830}{\frac{200}{\sqrt{25}}} = -3.250$
- $df = 24, t = -3.25 $
- Two Tail Test
- https://naneja.github.io/python/t-table
- $p = 0.0025 + 0.0025 = 0.005 < 0.05$

- Reject the null in favor of $H_A: \mu \ne 1830$

What is the Confidence Interval for the population for Company A?

95% Confidence Interval

For 95% CI, we need t-critical value at p=0.025 at df = 24

- $df=24, ~p=0.025 \implies t_{critical} = 2.064 $
- $\pm ~\text{t_critical} * \text{std_error} = t * \frac{S}{\sqrt{n}} $
- $2.064 * \frac{200}{\sqrt{25}} = 82.56$
- CI = $(1700 - 82.56, 1700 + 82.56) ~=~ (1617.44, 1782.56) $

Margin of Error = 82.56

If n = 100

What are t-critical values

- $t = \frac{\bar{X} - 1830}{\frac{S}{\sqrt{n}}} = \frac{1700 - 1830}{\frac{200}{\sqrt{100}}} = -6.50$
- $df = 24, t = 6.5 $
- https://naneja.github.io/python/t-table
- $p = 0.00 + 0.00 = 0.00 < 0.05$

Reject the null in favor of $H_A: \mu \ne 1830$

What is the Confidence Interval for the population for Company A? - 95% Confidence Interval

- $df = 99, p = 0.025 \implies t_{critical} = $ 1.984
- Margin of Error = $\pm ~\text{t_critical} * \text{std_error} = t_{critical} * \frac{S}{\sqrt{n}} $
- ME = $1.984 * \frac{200}{\sqrt{100}} = 39.68$
- Increase of sample size will reduce Margin of error
- CI = $(1700 - 39.68, 1700 + 39.68) ~=~ (1660.32, 1739.68) $

# Cohen’s d

- Standardized mean difference that measures the distance between means in standardized units
- $Cohen’s~d = \frac{\bar{X}-\mu}{S}$
- In above Rent Example
- $\mu = 1830$
- Sample: $n = 25, \bar{X} = 1700, S = 200$
- $d = \frac{\bar{X}-\mu}{S} = \frac{1700 - 1830}{200} = -0.65$

# Dependent Samples

Same subject takes the test twice

Within subject designs

- each subject is assigned two conditions in random order
in control but get treatment

two kinds of treatment

Every subject is given a Pre-Test and a Post-Test

- Growth over time (Longitudinal Study)
- Each subject at different points of time

xi yi Di = xi - yi x1 y1 D1 = x1-y1 x2 y2 D2 = x2-y2 x3 y3 D3 = x3-y3 - each subject is assigned two conditions in random order

# Example - Keyboards

- Errors in two design of keyboards (QWERTY and Alphabetical)
- https://naneja.github.io/datasets
- file = Keyboards.csv

- https://naneja.github.io/datasets
- Mean Error
- n = 25
- Querty Keyboard = 5.08 and
- Alphabetical Keyboard = 7.8

Are these differences significant?

- $n = 25$
$H_0: \mu_Q = \mu_A ~and~ H_A: \mu_Q \ne \mu_A$

- Also can say $\mu_Q - \mu_A = 0$

What is Point Estimate for $\mu_Q - \mu_A$

- $\mu_Q - \mu_A = 5.08 - 7.8 = -2.72$

What is S

- d = QWERTYerrors - Alphabeticalerrors
- std_dev = $S = s_d = \sqrt{\frac{\Sigma (d-\bar{d})^2}{N-1}}$
- 3.69

- What is t-Statistic when S = 3.69
- $t = \frac{\mu_Q - \mu_A}{\frac{S}{\sqrt{n}}} = \frac{5.08 - 7.8}{\frac{3.69}{\sqrt{25}}} = -3.69$

- What is p-value
- Two Tail Test
- $df=24, t = -3.69$
- https://naneja.github.io/python/t-table

- $ p = 0.0005 + 0.0005 = 0.001 < 0.05 $

- Two Tail Test
Reject the Null or Fail to reject Null

- Reject the Null

- Significant Less Error and we may say causal effect due to keyboard design
- 95% Confidence Interval
- What are t-Critical Values for $\alpha=0.05$
- $ df = 24, p = 0.025$
- $t_{critical} = \pm 2.064$

- point estimate$ = \mu_Q - \mu_A = 5.08 - 7.8 = -2.72$
- $CI = \text{point estimate} \pm t_{critical} * \text{std error}$
- $CI = -2.72 - 2.064 * \frac{3.69}{\sqrt{25}}, ~ -2.72 + 2.064 * \frac{3.69}{\sqrt{25}}$
- -4.24, -1.20
- Users will make fewer errors in the range of 4 to 1 on querty keyboard than alpha errors

- What are t-Critical Values for $\alpha=0.05$

# Advantages and Disadvantages- Dependent Samples

- Within-Subject design
- Two Conditions
- Longitudinal
- Pre-Test, Post-Test

- Advantages
- Controls for individual differences
- Use Fewer Subjects
- Cost-Effective
- Less Time-consuming
- Less Expensive

- Disadvantages
- Carry-over Effects
- Second measurement can be affected by first treatment
- Order may influence results

- Within-Subject design

# Independent Samples

- Between-Subject Designs
Experimental

Observational

$ t = \frac{\bar{X_1}-\bar{X_2}}{standard~error} $

Standard Deviation = $\sqrt{S_1^2 + S_2^2}$

- We can’t take differences since these are not the same subjects (independent samples)

Standard Error $ = \frac{S}{\sqrt{n}} = \frac{\sqrt{S_1^2 + S_2^2}}{\sqrt{n}} = \sqrt{\frac{S_1^2 + S_2^2}{n}} = \sqrt{\frac{S1^2}{n} + \frac{S2^2}{n}} = \sqrt{\frac{S1^2}{n_1} + \frac{S2^2}{n_2}}$

Degrees of Freedom $ df= (n_1-1) + (n_2-1) = n_1 + n_2 -2$

- $ t = \frac{(\bar{X_1}-\bar{X_2})}{SE}$

# Example - Food Prices

- https://naneja.github.io/datasets
- file = FoodPrices.csv

- Hypothesis
- $H_0 : \mu_1 = \mu_2$
- $H_A : \mu_1 \ne \mu_2$

- $n1 = 18,~ n2 = 14$
- $df = n1 + n2 - 2 = 30$
- Sample Mean and Std Dev
- $\bar{X_1} = 8.94,~ \bar{X_2} = 11.14$
- $S_1 = 2.65,~ S_2 = 2.18$

- $S = \sqrt{\frac{S1^2}{n_1} + \frac{S2^2}{n_2}} = \sqrt{\frac{2.65^2}{18} + \frac{2.18^2}{14}} =0.85$
- $t = \frac{\bar{X_1} - \bar{X_2}}{S} = \frac{8.94 - 11.14}{0.85} =2.58$ (Ignore negative sign)
- $df = 30,~ t = 2.58$
- $p = 0.01 + 0.01 = 0.02 < 0.05$
- Reject the Null
- Prices are significantly different for both areas

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