• Develop an understanding of inference for two proportions
• Know how to compute the confidence interval of the difference of two proportions
• Know how to conduct a hypothesis test for the difference of two proportions

They take two groups of bees raised under identical conditions:

• Control Group \(A\): \(200\) bees with no pesticide exposure, where \(172\) survived; \[p_A = \frac{172}{200} = 0.86\]
• Treatment Group \(B\): \(200\) bees exposed to the pesticide, where \(140\) survived; \[p_B = \frac{140}{200} = 0.70\]

Research questions:

• Is there a real difference of proportions between the groups of is it just by chance?
• Is the lower survival rate in the treatment group a real biological effect of the pesticide, or just due to sampling variability?

Parameter of interest:

• The difference of the true proportions of the treatment and control groups. \[p_B - p_A\]

Point estimate:

• Difference of proportion of the sampled treatment and control groups. \[\hat{p}_B - \hat{p}_A\]

Confidence Interval:

• We can answer this research question using a confidence interval.
• In general, confidence interval are written as \[\text{point estimate} \pm z^* \cdot \text{SE}.\]

Sampling distribution:

• Assuming CLT, the sampling distribution of the difference of two proportions is a normal distribution with center \(\hat{p}_B - \hat{p}_A\), which is the point estimate.
• Standard error (SE) of a difference of two proportions will be determined by assuming that we can approximate a binomial with a normal distribution as long CLT conditions hold.

CLT Conditions:

• Each sample is independent
• Identically distributed observations with fixed population parameters \(p_A\) and \(p_B\)
• Both population distributions have finite variances \(p_A(1-p_A)\) and \(p_B(1-p_B)\)
• At least \(10\) “success” and \(10\) “failure” for both groups \(A\) and \(B\)

Normal approximation:

• Difference for two sample proportions will be nearly normally distributed with: \[ \begin{aligned} \overline{x} & \approx \hat{p}_B - \hat{p}_A \\ s^2 & \approx \hat{p}_A(1-\hat{p}_A) + \hat{p}_B(1-\hat{p}_B) \end{aligned} \]

Standard error:

• The general formula for the standard error (assuming CLT) is \[SE = \frac{s}{\sqrt{n}}.\]
• So, for two proportions, the standard error is \[ \begin{aligned} SE_{\hat{p}} & = \frac{\sqrt{\hat{p}_A(1-\hat{p}_A)}}{\sqrt{n}} + \frac{\sqrt{\hat{p}_B(1-\hat{p}_B)}}{\sqrt{n}} \\ SE_{\hat{p}} & = \sqrt{\frac{\hat{p}_A \left( 1-\hat{p}_A \right)}{n_A} + \frac{\hat{p}_B \left( 1-\hat{p}_B \right)}{n_B}} \end{aligned} \]

Information given:

• Estimate using a \(90\)% confidence interval. This is a confidence level of \(0.90\).

• Given: \(n_A = 200\), \(\hat{p}_A = \frac{172}{200} = 0.86\), and \(n_B = 200\), \(\hat{p}_B = \frac{140}{200} = 0.70\). First check conditions:

  • Independence: The sample is random, and \(200\) for both groups. We can assume independence here since the population of interest are bees, which is assume to be in millions.
  • Success-failure: We have \(172\) “success” and \(28\) “failure” in group \(A\), and We have \(140\) “success” and \(60\) “failure” in group \(B\)

• The CLT conditions hold. So, we can use a normal approximation of the sampling distribution of \(\hat{p}_B - \hat{p}_A\).

Confidence interval:

• For a \(0.90\) confidence level, \(z^* = 1.645\).
• The interval is calculated as \[ \begin{aligned} \hat{p}_B - \hat{p}_A & \pm z^* \cdot SE_{\hat{p}_B - \hat{p}_A} \\ \hat{p}_B - \hat{p}_A & \pm z^* \cdot \sqrt{\frac{\hat{p}_A \left( 1-\hat{p}_A \right)}{n_A} + \frac{\hat{p}_B \left( 1-\hat{p}_B \right)}{n_B}} \\ 0.16 & \pm 1.645 \cdot \sqrt{\frac{0.70 \left( 1-0.70 \right)}{200} + \frac{0.86 \left( 1-0.86 \right)}{200}} \end{aligned} \]
• The \(90\)% confidence interval is \((0.093,0.227)\).

Using R:

z_star <- qnorm(0.90+((1-0.90)/2),0,1) # critical value
n_A <- 200 # group A sample size
n_B <- 200 # group B sample size
p_hat_A <- 140/n_A # group A sample proportion
p_hat_B <- 172/n_B # group B sample proportion
p_diff <- p_hat_B-p_hat_A # difference in sample proportions (point estimate)
SE_diff <- sqrt(((p_hat_A*(1-p_hat_A))/n_A) + ((p_hat_B*(1-p_hat_B))/n_B)) # standard error
cl_lb <- p_diff - z_star*SE_diff # upper bound
cl_ub <- p_diff + z_star*SE_diff # lower bound
c(cl_lb,cl_ub) # interval as an ordered list

## [1] 0.09314525 0.22685475

Confidence interval:

• We are \(90\)% confident that the true difference in proportions of the treatment and control groups is between \(0.093\) and \(0.227\).
• If there were no actual difference, then \(\hat{p}_B - \hat{p}_A = 0\), a null value. Since \(0\) difference is not within the \(90\)% confidence interval, then we can say that the difference is significant.

Interpretation:

• If we repeat this experiment multiple times and compute the point estimate and the confidence interval, \(90\)% of the intervals will contain the true difference in proportions \(p_B - p_A\).

CLT conditions:

• Each sample is independent
• Identically distributed observations with a fixed population parameters \(p_A\) and \(p_B\)
• Population distribution have finite variance \(p_A(1-p_A)\) and \(p_B(1-p_B)\)
• Success-failure outcomes is \(n_Ap_A \ge 10\), \(n_A(1-p_A) \ge 10\), \(n_Bp_B \ge 10\), and \(n_B(1-p_B) \ge 10\)

Sampling distribution of the point estimate:

Confidence interval:

\[\hat{p}_{diff} \pm z^* \cdot SE_{\hat{p}_{diff}}\]

• The difference in sample proportions (point estimate) is \[\hat{p}_{diff} = \hat{p}_B - \hat{p}_B\] where \(\hat{p}_A\) and \(\hat{p}_B\) are the sample proportions for groups \(A\) and \(B\), respectively.
• The samples sizes for groups \(A\) and \(B\) are \(n_A\) and \(n_B\) respectively.
• The standard error is \[SE_{\hat{p}_{diff}} = \sqrt{\frac{\hat{p}_A \left( 1-\hat{p}_A \right)}{n_A} + \frac{\hat{p}_B \left( 1-\hat{p}_B \right)}{n_B}}.\]
• The critical z-score for a given confidence level is \(z^*\).
• The margin of error is \(ME = z^* \cdot SE_{\hat{p}_{diff}}\).

Data:

• This study is an experiment where we try to determine whether pesticides reduces the chance of survival of honeybees.
• The point estimate is the differnce in sample proportions \(\hat{p}_B - \hat{p}_A = \frac{172}{200} - \frac{140}{200} = 0.16\) with sample sizes of \(n_A=200\) and \(n_B = 200\).
• Note that group \(B\) is the treatment group (with pesticide) and group \(A\) is the control group (no pesticide).

Objective:

• The sample difference is positive, meaning for a difference \(\hat{p}_B - \hat{p}_A > 0\), group \(B\) has the larger value.
• We need to use hypothesis testing to determine whether their is a significant difference in proportions between the treatment and control groups.

Null hypothesis \(H_0\): There is no difference in proportions between the treatment and control groups (there is no effect of the pesticide to the survival of the honeybees).

\[p_B - p_A = 0\]

Significance level: A significance level of \(\alpha = 0.10\) is chosen.

Alternative hypothesis \(H_A\): There is a difference in proportions between treatment and control groups (there is an effect of the pesticide to the survival of the honeybees).

\[p_B - p_A > 0\]

Test statistic for two proportions:

\[z = \frac{(\hat{p}_B - \hat{p}_A) - 0}{SE_{\hat{p}_B - \hat{p}_A}}\]

• The \(0\) term in the test statistic is the null value \(p_0\).
• \(SE_{\hat{p}_B - \hat{p}_A} = \sqrt{\hat{p}_{pool}\left( 1-\hat{p}_{pool} \right)\left( \frac{1}{n_A} + \frac{1}{n_B} \right)}\) is the pooled standard error.
• \(\hat{p}_{pool} = \frac{n_A \hat{p}_A + n_B \hat{p}_B}{n_A + n_B}\) is the pooled proportion.
• \(n_A\) and \(n_B\) are the sample sizes for group \(A\) and \(B\) respectively.

Computing the test statistic:

\[ \begin{aligned} \hat{p}_{pool} & = \frac{n_A \hat{p}_A + n_B \hat{p}_B}{n_A + n_B} \\ & = \frac{140 + 172}{200 + 200} \\ \hat{p}_{pool} & = 0.78 \end{aligned} \]

\[ \begin{aligned} z & = \frac{0.16}{\sqrt{0.78\left( 1-0.78 \right)\left( \frac{1}{200} + \frac{1}{200} \right)}} \\ z & \approx 3.862 \end{aligned} \]

Sampling distribution of the null value (normalized):

Using R:

n_A <- 200 # group A sample size
n_B <- 200 # group B sample size
p_A <- 140/n_A # group A sample proportion
p_B <- 172/n_B # group B sample proportion
p_pool <- (140+172)/(n_A+n_B) # pooled proportion
p_diff <- p_B - p_A # sample difference (point estimate)
p_0 <- 0 # null value
SE_pool <- sqrt(p_pool*(1-p_pool)*(1/n_A + 1/n_B)) # pooled standard error
z <- (p_diff-p_0)/SE_pool # test statistic

# p-value
1-pnorm(z,0,1) 

## [1] 5.61309e-05

Choices:

• If \(\text{p-value} < \alpha\), reject \(H_0\); there is enough evidence to support that there is an actual difference in proportions.
• If \(\text{p-value} \ge \alpha\), do not reject \(H_0\); there is not enough evidence to support that there is an actual difference in proportions.

Conclusions:

• We have a p-value of \(\approx 0\).
• Since \(0 < 0.10\), we reject \(H_0\).

Context:

• We conducted an experiment to see of pesticides have an effect ont eh survival of the bees.
• We grouped the beed into control (no pesticide) and treatment (with pesticide) groups and computed the proportion who survived.

Interpretation:

• Since we rejected the null, our claim that pesticides have an effect in survival is significant.
• There is enough evidence to support our claim.

Confidence Level:

• If we set a significance level \(\alpha = 0.10\), then the confidence level for the sample proportion is \(1-\alpha = 1 - 0.10 = 0.90\).
• The critical z-value of a \(0.90\) confidence level is \(z^* = 1.645\).

Confidence Interval:

• The \(90\)% confidence interval is \(0.16 \pm 1.645 \cdot 0.041\) or \((0.093,0.227)\).

State the Hypotheses:

• Null Hypothesis \(H_0\): The difference in population proportion remains unchanged. \[p_B - p_A = p_0\]
• Alternative Hypothesis \(H_A\): The difference in population proportion has changed. \[p_B - p_A \ne p_0\]

Set Significance Value \(\alpha\):

• Common values are \(\alpha = 0.10, 0.05, 0.01\).
• Note that \(\alpha\) is the Type I error rate.

Compute the test statistic:

\[z = \frac{\left(\hat{p}_B-\hat{p}_A\right)-p_0}{SE_{p_B - p_A}}\]

• \(\hat{p}_A\) is the point estimate for group \(A\)
• \(\hat{p}_B\) is the point estimate for group \(B\)
• \(SE_{\hat{p}_B - \hat{p}_A} = \sqrt{\hat{p}_{pool}\left( 1-\hat{p}_{pool} \right)\left( \frac{1}{n_A} + \frac{1}{n_B} \right)}\) is the standard error
• \(\hat{p}_{pool} = \frac{n_A \hat{p}_A + n_B \hat{p}_B}{n_A + n_B}\)

Determine the p-value:

• If one-sided test:

  • Find \(P(Z \le z | H_0)\) for left tail
  • Find \(1-P(Z \ge z | H_0)\) for right tail

• If two-sided test:

  • Find \(2 \cdot P(Z \le z | H_0)\) or \(2 \cdot (1-P(Z \ge z | H_0))\)

• Note that \(Z \sim N(0,1)\) is an r.v. with the standard normal distribution.

Sampling distribution of the null value (left one-tail):

Sampling distribution of the null value (right one-tail):

Sampling distribution of the null value (two-tail):

Make a decision and conclusion:

• Reject \(H_0\) if the \(\text{p-value} < \alpha\): There is enough evidence to support \(H_A\) that there is a significant difference in sample proportions under the null hypothesis.
• Fail to reject the \(H_0\) if the \(\text{p-value} \ge \alpha\): There is not enough evidence to support \(H_A\) that there is a significant difference in sample proportions under the null hypothesis.

Important Notes: