Data Science Knowledge Sharing Session: 3

Hypothesis Testing

Hypothesis testing is a statistical method used to determine the validity of a claim or hypothesis about a population based on sample data. It involves making an educated guess, or a hypothesis, about a characteristic or parameter of a population, and then using statistical analysis to assess the evidence in favor of or against the hypothesis.

The general process of hypothesis testing involves the following steps:

1.    Formulate a null hypothesis (H0): This is the default assumption or claim that is being tested. It is typically stated as there being no significant difference or relationship between variables.

2.    Formulate an alternative hypothesis (H1 or Ha): This is the opposite of the null hypothesis and represents the claim that the researcher is trying to support or establish.

3.    Select a significance level (α): This is the threshold used to determine whether the evidence is strong enough to reject the null hypothesis. Common significance levels are 0.05 (5%) and 0.01 (1%).

4.    Collect and analyze data: Data is collected from a sample, and statistical tests are performed to calculate a test statistic, such as a t-score or z-score, which quantifies the strength of the evidence against the null hypothesis.

5.    Calculate p-value: The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true. It is compared to the significance level to determine whether to reject or fail to reject the null hypothesis.

6.    Make a decision: If the p-value is less than the significance level, the null hypothesis is rejected in favor of the alternative hypothesis. If the p-value is greater than the significance level, the null hypothesis is not rejected.

7.    Interpret the results: The findings are interpreted in the context of the research question and conclusions are drawn accordingly.

Parametric Tests:

Parametric tests are statistical tests that make assumptions about the underlying probability distribution of the data, typically assuming that the data follows a normal distribution. These tests are commonly used when dealing with continuous data (interval or ratio scale) and are based on population parameters such as means, variances, and standard deviations. Parametric tests are generally considered more powerful than non-parametric tests when the assumptions are met, but they can be sensitive to violations of these assumptions.

Here are some commonly used parametric tests, along with their mathematical formulas and examples:

               I.         One-Sample t-test:

The one-sample t-test is a parametric statistical test used to compare the mean of a single sample to a known or hypothesized value. It is commonly used to determine if a sample mean is significantly different from a hypothesized population mean. The formula for the one-sample t-test is as follows:

Formula: t = (x̄ - μ) / (s / sqrt(n))

where:

t: test statistic

x̄: sample mean

μ: hypothesized population mean

s: sample standard deviation

n: sample size

Example: Let's say we have a sample of 30 students who took a math exam, and we want to determine if the average score of the sample is significantly different from the hypothesized population mean of 80. The sample mean is found to be 76, with a sample standard deviation of 5. The sample size is 30.

Step 1: Define hypotheses: The null hypothesis (H0): The sample mean is not significantly different from the hypothesized population mean. μ = 80 The alternative hypothesis (Ha): The sample mean is significantly different from the hypothesized population mean. μ ≠ 80

Step 2: Calculate the test statistic:

t = (x̄ - μ) / (s / sqrt(n))

t = (76 - 80) / (5 / sqrt (30))

t = -4 / (5 / 5.48)

t = -4 / 0.91

t = -4.40 (rounded to two decimal places)

Step 3: Determine the critical value or p-value: We can look up the critical value for a two-tailed test with a significance level of α (e.g., α = 0.05) in a t-distribution table or use statistical software to calculate the p-value associated with the t-statistic.

Step 4: Make a decision and interpret results: If the calculated t-statistic falls in the critical region (i.e., t-statistic > critical value or p-value < α), we reject the null hypothesis and conclude that there is a statistically significant difference between the sample mean and the hypothesized population mean. Otherwise, we fail to reject the null hypothesis.

For example, if the critical value for a two-tailed test with α = 0.05 is ±2.04, and our calculated t-statistic is -4.40, which falls in the critical region, we would reject the null hypothesis and conclude that there is a statistically significant difference between the sample mean and the hypothesized population mean at a significance level of 0.05.

             II.         Z-test

Z-test is another commonly used parametric test in hypothesis testing. The Z-test is used to compare a sample mean to a known or hypothesized population mean when the population standard deviation is known. It is often used when the sample size is large (typically n > 30) and the data follows a normal distribution.

The mathematical formula for the Z-test is:

Z = (X - μ) / (σ / √n)

where:

Z is the test statistic

X is the sample mean

μ is the population mean (the hypothesized value being tested)

σ is the population standard deviation

n is the sample size

The Z-test statistic is then compared to a critical value (e.g., from the standard normal distribution) or the p-value is calculated to determine the statistical significance of the test.

Example: Let's say we want to test if the average height of a sample of 100 adults is significantly different from a known population average height of 5 feet 6 inches (66 inches), with a known population standard deviation of 3 inches. We can use the Z-test to perform this hypothesis test.

Sample size (n) = 100

Sample mean (X) = 68 inches

Population mean (μ) = 66 inches

Population standard deviation (σ) = 3 inches

Using the Z-test formula, we can calculate the test statistic:

Z = (X - μ) / (σ / √n) Z =

(68 - 66) / (3 / √100) Z =

2 / 0.3 Z ≈ 6.67

Assuming a significance level (alpha) of 0.05, we can compare the calculated Z-test statistic to the critical value from the standard normal distribution (e.g., Z-critical value for a two-tailed test at alpha = 0.05 is approximately ±1.96) or calculate the p-value associated with the Z-test statistic.

If the calculated Z-test statistic exceeds the critical value or the p-value is less than the significance level (0.05), then we would reject the null hypothesis and conclude that there is a statistically significant difference between the sample mean and the hypothesized population mean. Otherwise, we would fail to reject the null hypothesis. 

           III.         Independent Samples t-test:

The independent samples t-test, also known as the two-sample t-test, is a parametric statistical test used to compare the means of two independent samples. It is commonly used to determine if there is a statistically significant difference between the means of two groups or conditions. The formula for the independent samples t-test is as follows:

Formula: t = (x̄1 - x̄2) / sqrt ((s1^2 / n1) + (s2^2 / n2))

where:

t: test statistic

x̄1, x̄2: sample means of Group 1 and Group 2, respectively

s1, s2: sample standard deviations of Group 1 and Group 2, respectively

n1, n2: sample sizes of Group 1 and Group 2, respectively

Example: Let's say we want to compare the average test scores of two groups of students, Group 1 and Group 2, to determine if there is a significant difference. Group 1 consists of 40 students with a sample mean of 78 and a sample standard deviation of 10. Group 2 consists of 35 students with a sample mean of 82 and a sample standard deviation of 12.

Step 1: Define hypotheses: The null hypothesis (H0): There is no significant difference between the means of Group 1 and Group 2. μ1 = μ2 the alternative hypothesis (Ha): There is a significant difference between the means of Group 1 and Group 2. μ1 ≠ μ2

Step 2: Calculate the test statistic:

t = (x̄1 - x̄2) / sqrt ((s1^2 / n1) + (s2^2 / n2))

t = (78 - 82) / sqrt ((10^2 / 40) + (12^2 / 35))

t = -4 / sqrt (2.5 + 4.73) t = -4 / sqrt (7.23)

t = -4 / 2.69 t = -1.49 (rounded to two decimal places)

Step 3: Determine the critical value or p-value: We can look up the critical value for a two-tailed test with a significance level of α (e.g., α = 0.05) in a t-distribution table or use statistical software to calculate the p-value associated with the t-statistic.

Step 4: Make a decision and interpret results: If the calculated t-statistic falls in the critical region (i.e., t-statistic > critical value or p-value < α), we reject the null hypothesis and conclude that there is a statistically significant difference between the means of Group 1 and Group 2. Otherwise, we fail to reject the null hypothesis.

For example, if the critical value for a two-tailed test with α = 0.05 is ±1.99, and our calculated t-statistic is -1.49, which does not fall in the critical region, we would fail to reject the null hypothesis and conclude that there is no statistically significant difference between the means of Group 1 and Group 2 at a significance level of 0.05.

           IV.         Paired Samples t-test:

The paired samples t-test, also known as the dependent samples t-test or matched pairs t-test, is a parametric statistical test used to compare the means of two related or paired samples. It is commonly used to determine if there is a statistically significant difference between the means of two groups or conditions where the data points are paired or matched in some way, such as before-and-after measurements, repeated measures, or matched pairs. The formula for the paired samples t-test is as follows:

Formula: t = (x̄d - μd) / (sᵈ / √n)

where:

t: test statistic

x̄d: sample mean of the differences between paired observations

μd: hypothesized mean difference (often assumed to be 0 under the null hypothesis)

sᵈ: sample standard deviation of the differences between paired observations

n: number of pairs or observations

Example: Let's say we want to determine if there is a statistically significant difference in the blood pressure readings of a group of patients before and after a treatment intervention. We collect blood pressure measurements from 30 patients before the treatment (pre-treatment) and after the treatment (post-treatment), and we want to determine if there is a significant difference in the means of these paired observations.

Step 1: Define hypotheses: The null hypothesis (H0): There is no significant difference in the means of the paired observations. μd = 0 The alternative hypothesis (Ha): There is a significant difference in the means of the paired observations. μd ≠ 0

Step 2: Calculate the test statistic: t = (x̄d - μd) / (sᵈ / √n) where x̄d is the sample mean of the differences, μd is the hypothesized mean difference (often assumed to be 0 under the null hypothesis), sᵈ is the sample standard deviation of the differences, and n is the number of pairs or observations.

Calculate the t-statistic: The t-statistic for a paired samples t-test is calculated as follows: t = (Mdiff - μdiff) / (SDdiff / sqrt(N))

where:

Mdiff: Mean of the differences between the paired samples (i.e., post-test scores minus pre-test scores)

μdiff: Expected mean of the differences under the null hypothesis (typically set to 0)

SDdiff: Standard deviation of the differences between the paired samples

N: Number of paired samples

Example: Let's say the mean of the differences (Mdiff) is 5.2, the expected mean of the differences under the null hypothesis (μdiff) is 0, the standard deviation of the differences (SDdiff) is 3.8, and the number of paired samples (N) is 20.

t = (5.2 - 0) / (3.8 / sqrt (20))

t = 5.2 / (3.8 / 4.47)

t = 5.2 / 0.849

t = 6.13 (rounded to two decimal places)

Determine the degrees of freedom: The degrees of freedom for a paired samples t-test is equal to the number of paired samples minus 1 (N-1). In our example, the degrees of freedom would be 20-1 = 19.

Step 3: Determine the critical value or p-value: We can look up the critical value for a two-tailed test with a significance level of α (e.g., α = 0.05) in a t-distribution table or use statistical software to calculate the p-value associated with the t-statistic.

Step 4: Make a decision and interpret results: If the calculated t-statistic falls in the critical region (i.e., t-statistic > critical value or p-value < α), we reject the null hypothesis and conclude that there is a statistically significant difference in the means of the paired observations. Otherwise, we fail to reject the null hypothesis.

Let's assume we have obtained a critical value of 2.093 (for a two-tailed test) and a p-value of 0.001 (from a statistical software or t-distribution table) for the calculated t-statistic of 6.13 with 19 degrees of freedom in the paired samples t-test example provided earlier.

Interpretation using critical value: If we are using a significance level of 0.05, the critical value for a two-tailed test with 19 degrees of freedom is 2.093. Since the calculated t-statistic of 6.13 is greater than the critical value of 2.093, we would reject the null hypothesis. This suggests that there is a statistically significant difference between the pre-test and post-test scores of the students after undergoing the intervention program at a significance level of 0.05.

Interpretation using p-value: If we are using a significance level of 0.05, and the obtained p-value is 0.001, which is less than the significance level, we would reject the null hypothesis. This suggests that there is a statistically significant difference between the pre-test and post-test scores of the students after undergoing the intervention program at a significance level of 0.05.

In both interpretations, we reject the null hypothesis and conclude that there is a statistically significant difference between the pre-test and post-test scores of the students after undergoing the intervention program.

             V.         Analysis of Variance (ANOVA):

ANOVA, or Analysis of Variance, is a parametric statistical test used to compare the means of three or more groups or conditions. ANOVA is used to determine if there is a statistically significant difference among the means of the groups being compared. It is a widely used test in many fields, such as psychology, biology, business, and social sciences, to name a few. The formula for one-way ANOVA is as follows:

Formula: F = (MS between) / (MS within)

where:

F: test statistic (F-ratio)

MS between: mean square between-groups (variance between groups divided by the degrees of freedom between groups)

MS within: mean square within-groups (variance within groups divided by the degrees of freedom within groups)

Example: Let's say we want to compare the mean scores of three different study groups (Group A, Group B, and Group C) to determine if there is a statistically significant difference in their test scores. We have collected data on the test scores of each group, with 10 participants in each group.

Step 1: Define hypotheses: The null hypothesis (H0): There is no significant difference among the means of the groups. The alternative hypothesis (Ha): There is a significant difference among the means of the groups.

Step 2: Calculate the test statistic: To calculate the F-ratio, we need to first calculate the sum of squares (SS) for between-groups and within-groups, and then divide these by the respective degrees of freedom (df) to get the mean squares (MS). Finally, we divide MS between by MS within to get the F-ratio.

calculating the test statistic (F-ratio) for a one-way ANOVA with three groups (Group A, Group B, and Group C) using some hypothetical data:

Group A: 8, 10, 12, 9, 11

Group B: 14, 16, 18, 15, 17

Group C: 20, 22, 24, 21, 23

Calculate the sum of squares (SS) for between-groups: SSbetween = ∑ (n * (mean of each group - grand mean) ^2)

where:

n: number of participants in each group

mean of each group: average score of each group

grand mean: average of all scores across all groups

For

Group A: n = 5 (since there are 5 participants in Group A)

mean of Group A = (8 + 10 + 12 + 9 + 11) / 5 = 10

grand mean = (8 + 10 + 12 + 9 + 11 + 14 + 16 + 18 + 15 + 17 + 20 + 22 + 24 + 21 + 23) / 15 ≈ 16.87 (average of all scores)

SSbetween_GroupA = 5 * (10 - 16.87) ^2 = 5 * (-6.87) ^2 ≈ 236.66

Similarly, we can calculate SS between for Group B and Group C.

For

Group B: n = 5

mean of Group B = (14 + 16 + 18 + 15 + 17) / 5 = 16

SSbetween_GroupA = 5 * (16 - 16.87) ^2 = 5 * (-0.87) ^2 ≈ 3.61

For

Group C: n = 5

mean of Group C = (20 + 22 + 24 + 21 + 23) / 5 = 22

SSbetween_GroupA = 5 * (22 - 16.87) ^2 = 5 * 5.13^2 ≈ 131.56

Calculate the sum of squares (SS) for within-groups: SS within = ∑ ((n - 1) * variance of each group)

where:

n: number of participants in each group

variance of each group: variance of scores within each group

For

Group A: n = 5

variance of Group A = [(8 - 10) ^2 + (10 - 10) ^2 + (12 - 10) ^2 + (9 - 10) ^2 + (11 - 10) ^2] / (5 - 1) = 4 / 4 = 1

SSwithin_GroupA = (5 - 1) * 1 = 4

Similarly, we can calculate SS within for Group B and Group C.

For

Group B: n = 5

variance of Group B = [(14 - 16) ^2 + (16 - 16) ^2 + (18 - 16) ^2 + (15 - 16) ^2 + (17 - 16) ^2] / (5 - 1) = 4 / 4 = 1

SSwithin_GroupA = (5 - 1) * 1 = 4

For

Group C: n = 5

variance of Group C = [(20 - 22) ^2 + (22 - 22) ^2 + (24 - 22) ^2 + (21 -22)^2 + (23 - 22) ^2] / (5 - 1) = 4 / 4 = 1

SSwithin_GroupA = (5 - 1) * 1 = 4

Calculate the F-ratio: F = MS between / MS within

where:

MS between: Mean Square between-groups, calculated as SS between / (k - 1), where k is the number of groups

MS within: Mean Square within-groups, calculated as SS within / (N - k), where N is the total number of participants and k is the number of groups

For our example: k = 3 (since we have 3 groups: Group A, Group B, Group C) N = 15 (total number of participants, 5 in each group)

MS between = (SSbetween_GroupA + SSbetween_GroupA + SSbetween_GroupA) / (k - 1) = (236.66 + 3.61 + 131.56) / 2 ≈ 185.92

MS within = (SSwithin_GroupA + SSwithin_GroupA + SSwithin_GroupA) / (N - k) = (4 + 4 + 4) / (15 - 3) ≈ 0.53

F = MS between / MS within = 185.92 / 0.53 ≈ 350.61

So, the test statistic (F-ratio) for this one-way ANOVA example is approximately 350.61.

Step 3: Determine the critical value or p-value: We can look up the critical value for an F-test with appropriate degrees of freedom and significance level (e.g., α = 0.05) in an F-distribution table or use statistical software to calculate the p-value associated with the F-ratio.

Step 4: Make a decision and interpret results: If the calculated F-ratio falls in the critical region (i.e., F-ratio > critical value or p-value < α), we reject the null hypothesis and conclude that there is a statistically significant difference among the means of the groups. Otherwise, we fail to reject the null hypothesis.

For the above example, Let’s assume that the chosen significance level (alpha) is set at 0.05 for both the critical value and p-value approaches.

Interpretation using critical value: The critical value for an F-distribution depends on the degrees of freedom of the numerator (k-1) and the degrees of freedom of the denominator (N-k), where k is the number of groups and N is the total number of participants. In our example, k = 3 (Group A, Group B, and Group C) and N = 15 (total number of participants).

If we consult an F-distribution table (or use statistical software), with degrees of freedom (2,12) for the numerator and denominator respectively (calculated as (k-1) and (N-k)), we find that the critical value at alpha = 0.05 is approximately 3.89.

Since our calculated F-ratio is 350.61, which is much larger than the critical value of 3.89, we would reject the null hypothesis. This means that there is a statistically significant difference between at least two of the group means in terms of their scores on the outcome variable.

Interpretation using p-value: The p-value associated with the calculated F-ratio of 350.61 can be obtained using statistical software or consulting an F-distribution table with degrees of freedom (2,12) for the numerator and denominator.

If the obtained p-value is less than the chosen significance level of 0.05, then we would reject the null hypothesis. In this case, if the p-value is very small (e.g., p < 0.05), it would indicate strong evidence against the null hypothesis and support the conclusion that there is a statistically significant difference between at least two of the group means.

Since our calculated p-value is likely to be much smaller than 0.05 (given the large calculated F-ratio of 350.61), we would reject the null hypothesis and conclude that there is a statistically significant difference between at least two of the group means in terms of their scores on the outcome variable.

           VI.         Pearson's Correlation Coefficient:

Pearson's correlation coefficient is a statistical measure that quantifies the strength and direction of a linear relationship between two continuous variables. In hypothesis testing, it is used to determine if there is a statistically significant correlation between two variables.

Formula: The formula for Pearson's correlation coefficient, denoted by the symbol "r", is as follows:

r = (Σ((Xi - X̄)(Yi - Ȳ))) / sqrt(Σ((Xi - X̄)^2) * Σ((Yi - Ȳ)^2))

where:

Xi and Yi are the individual values of the two variables being correlated

X̄ and Ȳ are the means of the two variables, respectively

Σ denotes the sum of the values across all the data points

Example: Let's say we have two variables, X and Y, and we want to test if there is a correlation between them. We have the following data:

X: 3, 5, 6, 7, 9

Y: 12, 15, 17, 20, 22

Step 1: Calculate the means of X and Y.

 X̄ = (3 + 5 + 6 + 7 + 9) / 5 = 6

Ȳ = (12 + 15 + 17 + 20 + 22) / 5 = 17.2

Step 2: Calculate the numerator of the correlation coefficient formula.

Σ ((Xi - X̄) (Yi - Ȳ)) =

(3 - 6) (12 - 17.2) + (5 - 6) (15 - 17.2) + (6 - 6) (17 - 17.2) + (7 - 6) (20 - 17.2) + (9 - 6) (22 - 17.2)

= (-3) (-5.2) + (-1) (-2.2) + (0) (-0.2) + (1) (2.8) + (3) (4.8)

= 15.6 + 2.2 + 0 + 2.8 + 14.4

= 35

Step 3: Calculate the denominators of the correlation coefficient formula.

Σ ((Xi - X̄) ^2) =

(3 - 6) ^2 + (5 - 6) ^2 + (6 - 6) ^2 + (7 - 6) ^2 + (9 - 6) ^2 =

9 + 1 + 0 + 1 + 9

= 20

Σ ((Yi - Ȳ) ^2) =

(12 - 17.2) ^2 + (15 - 17.2) ^2 + (17 - 17.2) ^2 + (20 - 17.2) ^2 + (22 - 17.2) ^2=

27.04 + 4.84 + 0.04 + 7.84 + 22.09 =

62.85

Step 4: Plug the values into the correlation coefficient formula.

r = (Σ ((Xi - X̄) (Yi - Ȳ))) / sqrt (Σ ((Xi - X̄) ^2) * Σ ((Yi - Ȳ) ^2)) =

35 / sqrt (20 * 62.85) =

35 / sqrt (1257)

Step 5: Interpret the results. Once we have calculated the value of r, we can compare it to a critical value or use a p-value to determine if the correlation is statistically

Interpretation using Critical Value: To interpret the results using a critical value, we would need to compare the calculated correlation coefficient (r) to the critical value at a given significance level (α).

For example, let's assume a significance level (α) of 0.05, which is commonly used in hypothesis testing. We can look up the critical value for a two-tailed test with a sample size of n-2 degrees of freedom (df) in a standard critical value table or use a calculator.

If the calculated correlation coefficient (r) is greater than the critical value, we would reject the null hypothesis and conclude that there is a statistically significant correlation between the two variables. If the calculated correlation coefficient (r) is less than the critical value, we would fail to reject the null hypothesis and conclude that there is not enough evidence to support a statistically significant correlation between the two variables.

Interpretation using p-value: To interpret the results using a p-value, we would need to compare the calculated p-value to the significance level (α).

If the p-value is less than the significance level (α), typically 0.05, we would reject the null hypothesis and conclude that there is a statistically significant correlation between the two variables. This suggests that the observed correlation is unlikely to have occurred by chance alone.

If the p-value is greater than the significance level (α), we would fail to reject the null hypothesis and conclude that there is not enough evidence to support a statistically significant correlation between the two variables.

It's important to note that the interpretation of results should be based on both the magnitude of the correlation coefficient (r) and the associated p-value or critical value, as well as the context of the study and the research question being investigated. It's also important to consider the sample size and potential limitations of the data when interpreting the results of a correlation analysis.

         VII.         Chi-Square Test of Independence:

The chi-square test of independence is a statistical test used to determine if there is a significant association between two categorical variables in a contingency table. The contingency table is a tabular representation of the joint frequencies of two categorical variables, with rows representing one variable and columns representing the other variable.

The mathematical formula for the chi-square test of independence is as follows:

Chi-square statistic (χ^2) = Σ [(Observed frequency - Expected frequency) ^2 / Expected frequency]

Where:

Σ denotes the sum of all the values calculated for each cell in the contingency table.

Observed frequency is the actual frequency observed in each cell of the contingency table.

Expected frequency is the frequency that would be expected if there was no association between the two variables, calculated based on the assumption of independence.

The expected frequency for each cell is calculated as follows: Expected frequency = (Row total * Column total) / Grand total

Once the chi-square statistic (χ^2) is calculated, it is compared to the critical value from the chi-square distribution with (r-1) (c-1) degrees of freedom, where r is the number of rows and c is the number of columns in the contingency table. If the calculated chi-square statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant association between the two categorical variables.

Here's an example to illustrate the chi-square test of independence:

Example: Suppose we want to determine if there is a significant association between gender (male, female) and smoking status (smoker, non-smoker) among a sample of 200 individuals. We collect data and construct a contingency table as follows:

               Smoking Status

           Smoker | Non-Smoker

Male | 40 | 60

Female | 20 | 80

We can use the chi-square test of independence to determine if there is a significant association between gender and smoking status.

Step 1: Set up hypotheses Null hypothesis (H0): There is no association between gender and smoking status. Alternative hypothesis (Ha): There is an association between gender and smoking status.

Step 2: Calculate the expected frequencies Expected frequency for each cell can be calculated using the formula: Expected frequency = (Row total * Column total) / Grand total

               Smoking Status

           Smoker | Non-Smoker

Male | 30 | 70

Female | 30 | 70

Step 3: Calculate the chi-square statistic Chi-square statistic (χ^2) = Σ [(Observed frequency - Expected frequency) ^2 / Expected frequency]

               Smoking Status

           Smoker | Non-Smoker

Male | (40-30) ^2/30 + (60-70) ^2/70 = 1.33

Female | (20-30) ^2/30 + (80-70) ^2/70 = 1.33

χ^2 = 1.33 + 1.33 = 2.66

Step 4: Determine critical value or p-value We can look up the critical value from the chi-square distribution table or use a statistical software to find the p-value associated with the calculated chi-square statistic.

Step 5: Make a decision If the calculated chi-square statistic (2.66) is greater than the critical value or the p-value is less than the significance level (e.g., α = 0.05), we would reject the null hypothesis and conclude that there is a significant association between gender and smoking status.

Interpretation: Based on the calculated chi-square statistic (2.66) and the p-value obtained from the chi-square distribution, let's say the p-value is 0.03, which is less than the significance level of 0.05. In this case, we would reject the null hypothesis and conclude that there is a significant association between gender and smoking status. This means that there is evidence to suggest that gender and smoking status are not independent of each other in the given sample. The association could be positive (indicating that one gender is more likely to be a smoker than the other) or negative (indicating the opposite). The direction and strength of the association can be further examined using additional statistical measures and techniques.

Note: As with any hypothesis test, it’s important to interpret the results in the context of the research question and consider the limitations and assumptions of the test. Additionally, reporting the effect size and confidence interval along with the p-value can provide a more comprehensive understanding of the results.