• Develop an understanding of inference for one proportion
• Know how to compute the confidence interval of one proportion
• Know how to conduct a hypothesis test for one proportion

Survey Question:

Which is the better way to test the drug?

• The first scientist wants to give the drug to 1000 people with high blood pressure and see how many of them experience lower blood pressure levels.
• The second scientist wants to give the drug to 500 people with high blood pressure, and not give the drug to another 500 people with high blood pressure, and see how many in both groups experience lower blood pressure levels.

Results:

• The GSS (General Social Survey) asks the same question, below is the distribution of responses from the 2010 survey:

Parameter of interest:

• Proportion of all Americans who have good intuition about experimental design. \[p \longrightarrow \text{a population proportion}\]

Point estimate:

• Proportion of sampled Americans who have good intuition about experimental design. \[\hat{p} \longrightarrow \text{a sample proportion}\]

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 sample proportion is a normal distribution with center \(\hat{p}\), which is the point estimate.
• Standard error (SE) of a sample proportion 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 a fixed population parameter \(p\)
• Population distribution have finite variance \(p(1-p)\)
• At least \(10\) “success” and \(10\) “failure”

Normal approximation:

• Sample proportions will be nearly normally distributed with: \[ \begin{aligned} \overline{x} & \approx \hat{p} \\ s^2 & \approx \hat{p}(1-\hat{p}) \end{aligned} \]

Standard error:

• The general formula for the standard error (assuming CLT) is \[SE = \frac{s}{\sqrt{n}}.\]
• So, for one proportion, the standard error is \[ \begin{aligned} SE_{\hat{p}} & = \frac{\sqrt{\hat{p}(1-\hat{p})}}{\sqrt{n}} \\ SE_{\hat{p}} & = \sqrt{\frac{\hat{p}(1-\hat{p}}{n}} \end{aligned} \]

Information given:

• Estimate using a \(95\)% confidence interval. This is a confidence level of \(0.95\).

• Given: \(n = 670\), \(\hat{p} = \frac{571}{670} \approx 0.852\). First check conditions:

  • Independence: The sample is random, and \(670\) which is less than \(1\)% of all Americans, therefore we can assume that one respondent’s response is independent of another.
  • Success-failure: \(571\) people answered correctly (“success”) and \(99\) answered incorrectly (“failure”), both are greater than \(10\).

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

Confidence interval:

• For a \(0.95\) confidence level, \(z^* \approx 1.96\).
• The interval is calculated as \[ \begin{aligned} \hat{p} & \pm z^* \cdot SE_{\hat{p}} \\ \hat{p} & \pm z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p}}{n}} \\ 0.852 & \pm 1.960 \cdot \sqrt{\frac{0.852(1-0.852}{670}} \end{aligned} \]
• The \(95\)% confidence interval is \((0.825,0.879)\).

Using R:

z_star <- qnorm(0.95+((1-0.95)/2),0,1) # critical value
n <- 670 # sample size
p_hat <- 571/n # sample proportion (point estimate)
SE_p <- sqrt((p_hat*(1-p_hat))/n) # standard error
cl_lb <- p_hat - z_star*SE_p # upper bound
cl_ub <- p_hat + z_star*SE_p # lower bound
c(cl_lb,cl_ub) # interval as an ordered list

## [1] 0.8253686 0.8791090

Sampling distribution of the point estimate:

Confidence interval:

• We are \(95\)% confident that the true proportion of Americans who answered the question correctly is between \(0.825\) and \(0.879\).

Interpretation:

• If we repeat this survey multiple times and compute the point estimate and the confidence interval, \(95\)% of the intervals will contain the true proportion \(p\).

Margin of error:

How many people should we sample in order to cut the margin of error of a \(95\)% confidence interval down to \(0.01\)?

• For a sample size of \(n=670\) the margin of error is \(ME = 1.96 \cdot \sqrt{\frac{0.852(1-0.852}{670}} \approx 0.027\).
• The goal is to compute \(n\) so that \(ME = 0.01\).

Computing the number of samples:

\[ \begin{aligned} 1.96 \cdot \sqrt{\frac{0.852(1-0.852)}{n}} & \le 0.01 \\ 1.96^2 \times \frac{0.852(1-0.852)}{n} & \le 0.01^2 \end{aligned} \]

\[ \begin{aligned} n & \ge \left(\frac{1.96}{0.01}\right)^2 \left(0.852(1-0.852)\right) \\ n & \ge 4844.104 \end{aligned} \]

CLT conditions:

• Each sample is independent
• Identically distributed observations with a fixed population parameter \(p\)
• Population distribution have finite variance \(p(1-p)\)
• Success-failure outcomes is \(n\hat{p} \ge 10\) and \(n(1-\hat{p}) \ge 10\).

Sampling distribution of the point estimate:

Confidence interval:

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

• The sample proportion (point estimate) is \[\hat{p} = \frac{x}{n}\] where \(x\) is the number of desired outcomes and \(n\) is the sample size.
• The standard error is \[SE_{\hat{p}} = \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}.\]
• The critical z-score for a given confidence level is \(z^*\).
• The margin of error is \(ME = z^* \cdot SE_{\hat{p}}\).

Data:

• The research firm randomly surveys \(200\) customers who visited Brew Haven in the past month. Among them, \(173\) reported being satisfied with the service.
• So, the point estimate is the sample proportion \(\hat{p} = \frac{173}{200} \approx 0.865\) with sample size of \(n=200\).

Objective:

• Even though the sample proportion \(\hat{p} > 0.85\), we need to determine if the observation just happened by chance.
• We need to use hypothesis testing to determine whether the claim that at least \(85\)% of customers are satisfied is supported by the survey data.

Null Hypothesis \(H_0\): The satisfaction rate is equal to \(85\)%.

\[p = 0.85\]

Significance Level: A significance level of \(\alpha = 0.05\) is chosen.

Alternative Hypothesis \(H_A\): The satisfaction rate is greater than \(85\)%.

\[p > 0.85\]

Test statistic for one proportion:

\[z = \frac{\hat{p} - p_0}{SE_{p}}\]

• \(p_0 = 0.85\) is claimed proportion or the null value
• \(SE_{p} = \sqrt{\frac{p_0(1-p_0)}{n}}\) is the standard error of the null value’s sampling distribution
• \(n\) is the sample size

Computing the test statistic:

\[ \begin{aligned} z & = \frac{0.865 - 0.85}{\sqrt{\frac{0.85(1-0.85)}{200}}} \\ z & \approx 0.594 \end{aligned} \]

Sampling distribution of the null value:

Using R:

p_hat <- 173/200 # sample proportion (point estimate)
p_0 <- 0.85 # null value
n <- 200 # sample size
SE_p <- sqrt((p_0*(1-p_0))/(n)) # standard error
z <- (p_hat-p_0)/SE_p # test statistic

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

## [1] 0.2762265

Choices:

• If \(\text{p-value} < \alpha\), reject \(H_0\); there is enough evidence to support that the satisfaction rate is greater than \(85\)%.
• If \(\text{p-value} \ge \alpha\), do not reject \(H_0\); there is not enough evidence to support that the satisfaction rate is greater than \(85\)%.

Conclusions:

• We have a p-value of \(0.276\).
• Since \(0.276 > 0.05\), we failed to reject \(H_0\).

Context:

• The shop’s management claims that at least \(85\)% of its customers are satisfied.
• A market research firm is hired to assess their claim by conducting a survey.
• The population proportion \(p\) is the unknown true proportion of customers who are satisfied with the shop’s service.

Interpretation:

• Since we failed to reject the null, the shop’s management claim is not supported by the survey.
• There is not enough evidence to support the claim even though the sample proportion \(\hat{p} = 0.865\) is greater that the null value \(p_0 = 0.85\).

Confidence Level:

• If we set a significance level \(\alpha = 0.05\), then the confidence level for the sample proportion is \(1-\alpha = 1 - 0.05 = 0.95\).
• The critical z-value of a \(0.95\) confidence level is \(z^* = 1.96\).
• Standard error of the sample proportion is \(SE_{\hat{p}} = \sqrt{\frac{0.865(1-0.865)}{200}} \approx 0.024\).

Confidence Interval:

• The \(95\)% confidence interval is \(0.865 \pm 1.96 \times 0.024\) or \((0.818,0.912)\).

State the Hypotheses:

• Null Hypothesis \(H_0\): The population proportion remains unchanged. \[p = p_0\]
• Alternative Hypothesis \(H_A\): The population proportion has changed. \[p \ne p_0\]

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:

\[z = \frac{\hat{p}-p_0}{SE_p}\]

• \(\hat{p}\) is the point estimate
• \(SE_p = \sqrt{\frac{p_0\left( 1-p_0 \right)}{n}}\) is the standard error of the null value’s sampling distribution

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 the sample proportion is at least as extreme as the ones observed 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 the sample proportion is at least as extreme as the ones observed under the null hypothesis.

Important Notes: