1 Test Statistic for One Proportion Test in R

For \(\chi^2_1\), which is the chi-squared distribution with 1 degree of freedom, and \(z\), which is the standard normal distribution (mean is 0, and standard deviation is 1), the one proportion test has test statistics, \(\chi^2_1\) (or \(z^2\)), without continuity correction (correct = FALSE), which takes the form:

\[z = \frac{\frac{x}{n} - p_0}{\sqrt \frac{p_0(1-p_0)}{n}} \quad \text{or} \quad \chi^2_1=\sum_{i=1}^{2}\frac{(O_i-E_i)^2}{E_i}.\]

In cases where Yates’ continuity correction is applied (default in R), it takes the form:

\[z = \frac{\frac{(x \pm c)}{n} - p_0}{\sqrt \frac{p_0(1-p_0)}{n}} \quad \text{or} \quad \chi^2_1=\sum_{i=1}^{2}\frac{(|O_i-E_i|-c)^2}{E_i}.\] Note that \(z^2 = \chi^2_1\) in both cases. However, for one-sided tests, the level of significance \(\alpha\) for \(z\) is \(2\alpha\) for \(\chi^2_1\).

The test is ideal for large samples sizes (for example, \(np_0 > 5\) \(\&\) \(n(1-p_0) > 5\)).

See the exact binomial test for exact p-values. See also the two proportions test.

2 Simple One Proportion Test in R

With number of successes, \(x = 55\) in \(n = 100\) trials.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.05\), applying continuity correction.

\(H_0:\) the population proportion is equal to 0.5 (\(p = 0.5\)).

\(H_1:\) the population proportion is not equal to 0.5 (\(p \neq 0.5\), hence the default two-sided).

Because the level of significance is \(\alpha=0.05\), the level of confidence is \(1 - \alpha = 0.95\).

The prop.test() function has the default proportion as 0.5, the default alternative as "two.sided", the default level of confidence as 0.95, and the default method as continuity corrected, hence, you do not need to specify the "p", "alternative", "conf.level" and "correct" arguments in this case.

prop.test(55, 100, p = 0.5,
          alternative = "two.sided",
          conf.level = 0.95,
          correct = TRUE)

Or:

prop.test(55, 100)

    1-sample proportions test with continuity correction

data:  55 out of 100, null probability 0.5
X-squared = 0.81, df = 1, p-value = 0.3681
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.4475426 0.6485719
sample estimates:
   p 
0.55 

The sample proportion, \(\hat p = x/n\), is 0.55,

test statistic, \(\chi^2_1\), is 0.81,

the degree of freedom is 1,

the p-value, \(p\), is 0.3681,

the 95% confidence interval is [0.4475426, 0.6485719].

Interpretation:

Note that for prop.test() in R, the one proportion test’s p-value and T-statistic methods may disagree with the confidence interval method for some edge cases. This is because the p-value and T-statistic are based on the chi-squared (or equivalent standard normal) distribution, while the confidence interval is based on the Wilson Score method.

• P-value: With the p-value (\(p = 0.3681\)) being greater than the level of significance 0.05, we fail to reject the null hypothesis that the population proportion is equal to 0.5.

• \(\chi^2_1\) T-statistic: With test statistics value (\(\chi^2_1 = 0.81\)) being less than the critical value, \(\chi^2_{1,\alpha}=\text{qchisq(0.95, 1)}=3.8414588\) (or not in the shaded region), we fail to reject the null hypothesis that the population proportion is equal to 0.5.

• \(z\) T-statistic: With test statistics value (\(z=0.9=\sqrt{\chi^2_1 = 0.81}\)) being between than the critical values, \(-z_{\alpha/2}=\text{qnorm(0.025)}=-1.959964\) \(=-\sqrt{\text{qchisq(0.95, 1)}=3.8414588}\) and \(z_{\alpha/2}=\text{qnorm(0.975)}=1.959964\) \(=\sqrt{\text{qchisq(0.95, 1)}=3.8414588}\) (or not in the shaded region), we fail to reject the null hypothesis that the population proportion is equal to 0.5.

• Confidence Interval: With the null hypothesis proportion value (\(p = 0.5\)) being inside the confidence interval, \([0.4475426, 0.6485719]\), we fail to reject the null hypothesis that the population proportion is equal to 0.5.

par(mfrow=c(1,2))
#Plot Chi-squared
x = seq(0.01, 8, 1/1000); y = dchisq(x, df=1)
plot(x, y, type = "l",
     xlim = c(0, 8), ylim = c(-0.1, min(max(y), 1)),
     main = "One Proportion Test
Shaded Region for Simple Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.95, 1)
polygon(x = c(x[x >= point], 8, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.05"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(2, 0.4, 0.81, 0)
text(2, 0.45, "chi-squared = 0.81")
text(3.841459, -0.06, expression(chi[1][','][alpha]^2==3.841459))
#Plot z
x = seq(-4, 4, 1/1000); y = dnorm(x)
plot(x, y, type = "l",
     xlim = c(-4, 4), ylim = c(-0.03, max(y)),
     main = "One Proportion Test
Shaded Region for Simple Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point1 = qnorm(0.025); point2 = qnorm(0.975)
polygon(x = c(-4, x[x <= point1], point1),
        y = c(0, y[x <= point1], 0),
        col = "blue")
polygon(x = c(x[x >= point2], 4, point2),
        y = c(y[x >= point2], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.05"),
       fill = c("blue"), inset = 0.01)
# Add critical value and z-value
arrows(0, 0.1, 0.9, 0)
text(0, 0.12, expression(z==0.9))
text(-1.959964, -0.02, expression(-z[alpha/2]==-1.959964))
text(1.959964, -0.02, expression(z[alpha/2]==1.959964))

One Proportion Test Shaded Region for Simple Test in R

See line charts, shading areas under a curve, lines & arrows on plots, mathematical expressions on plots, and legends on plots for more details on making the plots above.

3 One Proportion Test Critical Value in R

To get the critical value for a one proportion test in R, you can use the qchisq() function for chi-squared distribution, or the qnorm() function for the standard normal distribution to derive the quantile associated with the given level of significance value \(\alpha\).

For two-tailed test with level of significance \(\alpha\). The critical values are: for chi-squared, qchisq(\(1-\alpha\), 1), or for \(z\), qnorm(\(\alpha/2\)) and qnorm(\(1-\alpha/2\)).

For one-tailed test with level of significance \(\alpha\). The critical value is: for chi-squared, qchisq(\(1-2\alpha\), 1); or for \(z\), left-tailed, qnorm(\(\alpha\)), and for right-tailed, qnorm(\(1-\alpha\)).

Example:

For \(\alpha = 0.05\).

Two-tailed:

#chi-squared
qchisq(0.95, 1)

[1] 3.841459

#z
qnorm(0.025); qnorm(0.975)

[1] -1.959964

[1] 1.959964

One-tailed:

#chi-squared
qchisq(0.9, 1)

[1] 2.705543

#Left z
qnorm(0.05)

[1] -1.644854

#Right z
qnorm(0.95)

[1] 1.644854

4 Two-tailed One Proportion Test in R

With number of successes, \(x = 31\) in \(n = 100\) trials.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.1\), without continuity correction.

\(H_0:\) the population proportion is equal to 0.4 (\(p = 0.4\)).

\(H_1:\) the population proportion is not equal to 0.4 (\(p \neq 0.4\), hence the default two-sided).

Because the level of significance is \(\alpha=0.1\), the level of confidence is \(1 - \alpha = 0.9\).

prop.test(31, 100, p = 0.4,
          alternative = "two.sided",
          conf.level = 0.9,
          correct = FALSE)

    1-sample proportions test without continuity correction

data:  31 out of 100, null probability 0.4
X-squared = 3.375, df = 1, p-value = 0.06619
alternative hypothesis: true p is not equal to 0.4
90 percent confidence interval:
 0.2397737 0.3902365
sample estimates:
   p 
0.31 

Interpretation:

• P-value: With the p-value (\(p = 0.06619\)) being lower than the level of significance 0.1, we reject the null hypothesis that the population proportion is equal to 0.4.

• \(\chi^2_1\) T-statistic: With test statistics value (\(\chi^2_1 = 3.375\)) being in the critical region (shaded area), that is, \(\chi^2_1 = 3.375\) greater than \(\chi^2_{1, \alpha}=\text{qchisq(0.9, 1)}=2.7055435\), we reject the null hypothesis that the population proportion is equal to 0.4.

• \(z\) T-statistic: With test statistics value (\(z=-1.8371173=-\sqrt{\chi^2_1 = 3.375}\)) being in the critical region (shaded area), that is, \(z=-1.8371173\) less than \(-z_{\alpha/2}=\text{qnorm(0.05)}=-1.6448536\) \(=-\sqrt{\text{qchisq(0.9, 1)}=2.7055435}\), we reject the null hypothesis that the population proportion is equal to 0.4.

• Confidence Interval: With the null hypothesis proportion value (\(p = 0.4\)) being outside the confidence interval, \([0.2397737, 0.3902365]\), we reject the null hypothesis that the population proportion is equal to 0.4.

par(mfrow=c(1,2))
#Plot Chi-squared
x = seq(0.01, 8, 1/1000); y = dchisq(x, df=1)
plot(x, y, type = "l",
     xlim = c(0, 8), ylim = c(-0.1, min(max(y), 1)),
     main = "One Proportion Test
Shaded Region for Two-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.9, 1)
polygon(x = c(x[x >= point], 8, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(3.375, 0.4, 3.375, 0)
text(3.375, 0.45, "chi-squared = 3.375")
text(2.705543, -0.06, expression(chi[1][','][alpha]^2==2.705543))
#Plot z
x = seq(-4, 4, 1/1000); y = dnorm(x)
plot(x, y, type = "l",
     xlim = c(-4, 4), ylim = c(-0.03, max(y)),
     main = "One Proportion Test
Shaded Region for Two-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point1 = qnorm(0.05); point2 = qnorm(0.95)
polygon(x = c(-4, x[x <= point1], point1),
        y = c(0, y[x <= point1], 0),
        col = "blue")
polygon(x = c(x[x >= point2], 4, point2),
        y = c(y[x >= point2], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and z-value
arrows(0, 0.15, -1.837117, 0)
text(0, 0.17, expression(z==-1.837117))
text(-1.644854, -0.02, expression(-z[alpha/2]==-1.644854))
text(1.644854, -0.02, expression(z[alpha/2]==1.644854))

One Proportion Test Shaded Region for Two-tailed Test in R

5 One-tailed One Proportion Test in R

Right Tailed Test

With number of successes, \(x = 92\) in \(n = 120\) trials.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.1\), applying continuity correction.

\(H_0:\) the population proportion is equal to 0.65 (\(p = 0.65\)).

\(H_1:\) the population proportion is greater than 0.65 (\(p > 0.65\), hence one-sided).

Because the level of significance is \(\alpha=0.1\), the level of confidence is \(1 - \alpha = 0.9\).

prop.test(92, 120, p = 0.65,
          alternative = "greater",
          conf.level = 0.9,
          correct = TRUE)

    1-sample proportions test with continuity correction

data:  92 out of 120, null probability 0.65
X-squared = 6.6758, df = 1, p-value = 0.004886
alternative hypothesis: true p is greater than 0.65
90 percent confidence interval:
 0.7093813 1.0000000
sample estimates:
        p 
0.7666667 

Interpretation:

• P-value: With the p-value (\(p = 0.004886\)) being less than the level of significance 0.1, we reject the null hypothesis that the population proportion is equal to 0.65.

• \(\chi^2_1\) T-statistic: With test statistics value (\(\chi^2_1 = 6.6758\)) being in the critical region (shaded area), that is, \(\chi^2_1 = 6.6758\) greater than \(\chi^2_{1,2\alpha}=\text{qchisq(0.8, 1)}=1.6423744\), we reject the null hypothesis that the population proportion is equal to 0.65.

• \(z\) T-statistic: With test statistics value (\(z=2.583757=\sqrt{\chi^2_1 = 6.6758}\)) being in the critical region (shaded area), that is, \(z=2.583757\) greater than \(z_{\alpha}=\text{qnorm(0.9)}=1.2815516\) \(=\sqrt{\text{qchisq(0.8, 1)}=1.6423744}\), we reject the null hypothesis that the population proportion is equal to 0.65.

• Confidence Interval: With the null hypothesis proportion value (\(p = 0.65\)) being outside the confidence interval, \([0.7093813, 1.0)\), we reject the null hypothesis that the population proportion is equal to 0.65.

par(mfrow=c(1,2))
#Plot Chi-squared
x = seq(0.01, 8, 1/1000); y = dchisq(x, df=1)
plot(x, y, type = "l",
     xlim = c(0, 8), ylim = c(-0.1, min(max(y), 1)),
     main = "One Proportion Test
Shaded Region for Right-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.8, 1)
polygon(x = c(x[x >= point], 8, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.2"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(6.6758, 0.4, 6.6758, 0)
text(6.6758, 0.45, "chi-squared = 6.6758")
text(1.642374, -0.06, expression(chi[1][','][2*alpha]^2==1.642374))
#Plot z
x = seq(-4, 4, 1/1000); y = dnorm(x)
plot(x, y, type = "l",
     xlim = c(-4, 4), ylim = c(-0.03, max(y)),
     main = "One Proportion Test
Shaded Region for Right-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qnorm(0.9)
polygon(x = c(x[x >= point], 4, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and z-value
arrows(2.583757, 0.15, 2.583757, 0)
text(2.583757, 0.17, expression(z==2.583757))
text(1.281552, -0.02, expression(z[alpha]==1.281552))

One Proportion Test Shaded Region for Right-tailed Test in R

Left Tailed Test

With number of successes, \(x = 24\) in \(n = 80\) trials.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.05\), without continuity correction.

\(H_0:\) the population proportion is equal to 0.35 (\(p=0.35\)).

\(H_1:\) the population proportion is less than 0.35 (\(p < 0.35\), hence one-sided).

Because the level of significance is \(\alpha=0.05\), the level of confidence is \(1 - \alpha = 0.95\).

prop.test(24, 80, p = 0.35,
          alternative = "less",
          conf.level = 0.95,
          correct = FALSE)

    1-sample proportions test without continuity correction

data:  24 out of 80, null probability 0.35
X-squared = 0.87912, df = 1, p-value = 0.1742
alternative hypothesis: true p is less than 0.35
95 percent confidence interval:
 0.0000000 0.3896842
sample estimates:
  p 
0.3 

Interpretation:

• P-value: With the p-value (\(p = 0.1742\)) being greater than the level of significance 0.05, we fail to reject the null hypothesis that the population proportion is equal to 0.35.

• \(\chi^2_1\) T-statistic: With test statistics value (\(\chi^2_1 = 0.87912\)) being less than the critical value, \(\chi^2_{1,2\alpha}=\text{qchisq(0.9, 1)}=2.7055435\) (or not in the shaded region), we fail to reject the null hypothesis that the population proportion is equal to 0.35.

• \(z\) T-statistic: With test statistics value (\(z=-0.937614=-\sqrt{\chi^2_1 = 0.87912}\)) being greater than the critical value, \(-z_{\alpha}=\text{qnorm(0.05)}=-1.6448536\) \(=-\sqrt{\text{qchisq(0.9, 1)}=2.7055435}\) (or not in the shaded region), we fail to reject the null hypothesis that the population proportion is equal to 0.35.

• Confidence Interval: With the null hypothesis proportion value (\(p = 0.35\)) being inside the confidence interval, \((0, 0.3896842]\), we fail reject the null hypothesis that the population proportion is equal to 0.35.

par(mfrow=c(1,2))
#Plot Chi-squared
x = seq(0.01, 8, 1/1000); y = dchisq(x, df=1)
plot(x, y, type = "l",
     xlim = c(0, 8), ylim = c(-0.1, min(max(y), 1)),
     main = "One Proportion Test
Shaded Region for Left-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.9, 1)
polygon(x = c(x[x >= point], 8, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(2.5, 0.4, 0.87912, 0)
text(2.5, 0.45, "chi-squared = 0.87912")
text(2.705543, -0.06, expression(chi[1][','][2*alpha]^2==2.705543))
#Plot z
x = seq(-4, 4, 1/1000); y = dnorm(x)
plot(x, y, type = "l",
     xlim = c(-4, 4), ylim = c(-0.03, max(y)),
     main = "One Proportion Test
Shaded Region for Left-tailed Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qnorm(0.05)
polygon(x = c(-4, x[x <= point], point),
        y = c(0, y[x <= point], 0),
        col = "blue")
legend("topright", c("Area = 0.05"),
       fill = c("blue"), inset = 0.01)
# Add critical value and z-value
arrows(0, 0.15, -0.937614, 0)
text(0, 0.17, expression(z==-0.937614))
text(-1.644854, -0.02, expression(-z[alpha]==-1.644854))

One Proportion Test Shaded Region for Left-tailed Test in R

6 One Proportion Test: Test Statistics, P-value & Degree of Freedom in R

Here for a one proportion test, we show how to get the test statistics (or chi-squared value), p-values, and degrees of freedom from the prop.test() function in R, or by written code.

x = 40; n = 65
prop_object = prop.test(x, n, p = 0.55,
                        alternative = "two.sided",
                        conf.level = 0.95,
                        correct = TRUE)
prop_object

    1-sample proportions test with continuity correction

data:  x out of n, null probability 0.55
X-squared = 0.87413, df = 1, p-value = 0.3498
alternative hypothesis: true p is not equal to 0.55
95 percent confidence interval:
 0.4861869 0.7308924
sample estimates:
        p 
0.6153846 

To get the test statistic or chi-squared value:

\[z = \frac{\frac{x}{n} - p_0}{\sqrt \frac{p_0(1-p_0)}{n}} \quad \text{or} \quad \chi^2_1=\sum_{i=1}^{2}\frac{(O_i-E_i)^2}{E_i}.\]

Applying Yates’ continuity correction, it takes the form:

\[z = \frac{\frac{(x \pm c)}{n} - p_0}{\sqrt \frac{p_0(1-p_0)}{n}} \quad \text{or} \quad \chi^2_1=\sum_{i=1}^{2}\frac{(|O_i-E_i|-c)^2}{E_i}.\]

\(c = \min\{0.5, |np_0-x|\}\), for \(z\) test statistic, \(x+c\) when \(np_0 \geq x\), and \(x-c\) when \(np_0<x\).

prop_object$statistic

X-squared 
0.8741259 

# to remove name X-squared
unname(prop_object$statistic)

[1] 0.8741259

Same as (with continuity correction):

x = 40; n = 65; p = 0.55
c = min(0.5, abs(n*p-x))
cc = c*sign(n*p-x)
#z
z = ((x+cc)/n - p)/sqrt(p*(1-p)/n)
z

[1] 0.934947

z^2 # convert to chi-squared

[1] 0.8741259

#chi-squared
chi = ((abs(x-n*p)-c)^2)/(n*p) + ((abs((n-x)-n*(1-p))-c)^2)/(n*(1-p))
chi

[1] 0.8741259

Without continuity correction:

#z
z = (x/n - p)/sqrt(p*(1-p)/n)
z
z^2 # convert to chi-squared
#chi-squared
chi = (abs(x-n*p)^2)/(n*p) + (abs((n-x)-n*(1-p))^2)/(n*(1-p))
chi

To get the p-value:

Two-tailed: With \(z = \sqrt{\chi^2_1}\): for positive test statistic (\(z^+\)), and negative test statistic (\(z^-\)).

For \(\chi^2_1\), \(P \left(\chi^2_1> \text{observed} \right)\).

For \(z\), \(Pvalue = 2*P(Z>z^+)\) or \(Pvalue = 2*P(Z<z^-)\).

One-tailed: For \(\chi^2_1\) is \(\frac{1}{2}*P \left(\chi^2_1> \text{observed} \right)\)

For \(z\), for right-tail, \(Pvalue = P(Z>z)\) or for left-tail, \(Pvalue = P(Z<z)\).

prop_object$p.value

[1] 0.3498156

Same as:

Note that the p-value depends on the \(\text{test statistics}\) (\(\chi^2_1 = 0.8741259\)), \(\text{degrees of freedom}\) (1), and \(z = \sqrt{\chi^2_1} = 0.934947\). We also use the distribution functions pchisq() for the chi-squared distribution and pnorm() for the standard normal distribution in R.

1-pchisq(0.8741259, 1)

[1] 0.3498156

2*(1-pnorm(sqrt(0.8741259))); 2*pnorm(-sqrt(0.8741259))

[1] 0.3498156

[1] 0.3498156

To get the degrees of freedom:

The degree of freedom is \(1\).

prop_object$parameter

df 
 1 

# to remove name df
unname(prop_object$parameter)

[1] 1

7 One Proportion Test: Estimate & Confidence Interval in R

Here for a one proportion test, we show how to get the sample proportion and confidence interval from the prop.test() function in R, or by written code.

x = 40; n = 60
prop_object = prop.test(x, n, p = 0.6,
                        alternative = "two.sided",
                        conf.level = 0.95,
                        correct = TRUE)
prop_object

    1-sample proportions test with continuity correction

data:  x out of n, null probability 0.6
X-squared = 0.85069, df = 1, p-value = 0.3564
alternative hypothesis: true p is not equal to 0.6
95 percent confidence interval:
 0.5320789 0.7798436
sample estimates:
        p 
0.6666667 

To get the point estimate or sample proportion:

\[\hat p = \frac{x}{n}.\]

prop_object$estimate

        p 
0.6666667 

# to remove name p
unname(prop_object$estimate)

[1] 0.6666667

Same as:

x = 40; n = 60; x/n

[1] 0.6666667

To get the (Wilson Score) confidence interval for \(p\):

R uses the Wilson Score confidence interval.

With standard error estimate: \[\widehat{\text{SE}} = \sqrt {\frac {\hat p(1- \hat p)}{n}},\]

and \(c = 0.5\) with continuity correction, \(c=0\) without continuity correction.

For two-tailed:

With \(\hat p = \frac{x-c}{n}\), the lower bound is:

\[\left( \frac{n}{n + z_{\alpha/2}^2}\right)\left\{\left(\widehat{p} + \frac{z_{\alpha/2}^2}{2n}\right) - z_{\alpha/2}\sqrt{ \widehat{\text{SE}}^2 + \frac{z_{\alpha/2}^2}{4n^2}}\right\}.\] With \(\hat p = \frac{x+c}{n}\), the upper bound is:

\[\left( \frac{n}{n + z_{\alpha/2}^2}\right)\left\{\left(\widehat{p} + \frac{z_{\alpha/2}^2}{2n}\right) + z_{\alpha/2}\sqrt{ \widehat{\text{SE}}^2 + \frac{z_{\alpha/2}^2}{4n^2}}\right\}.\]

For right one-tailed with \(\hat p = \frac{x-c}{n}\), the interval is:

\[ \left[\left( \frac{n}{n + z_{\alpha}^2}\right)*\left\{\left(\widehat{p} + \frac{z_{\alpha}^2}{2n}\right) - z_{\alpha}\sqrt{ \widehat{\text{SE}}^2 + \frac{z_{\alpha}^2}{4n^2}}\right\} \;,\; 1 \right].\]

For left one-tailed with \(\hat p = \frac{x+c}{n}\), the interval is:

\[ \left[ 0 \;,\; \left\{\left(\widehat{p} + \frac{z_{\alpha}^2}{2n}\right) + z_{\alpha}\sqrt{ \widehat{\text{SE}}^2 + \frac{z_{\alpha}^2}{4n^2}} \right\}*\left( \frac{n}{n + z_{\alpha}^2}\right) \right].\]

prop_object$conf.int

[1] 0.5320789 0.7798436
attr(,"conf.level")
[1] 0.95

# to remove attr conf.int
prop_object$conf.int[1:2]

[1] 0.5320789 0.7798436

Same as:

x = 40; n = 60; alpha = 0.05
crit = qnorm(1-alpha/2)
# Continuity, set c = 0 for no continuity correction
c = 0.5
# Set scale
mult = (n/(n + crit^2))
# Lower
phat = (x-c)/n; SE = sqrt(phat*(1-phat)/n)
p_cent = phat + crit^2/(2*n)
margin = crit*sqrt(SE^2+crit^2/(4*n^2))
l = mult*(p_cent - margin)
# Upper
phat = (x+c)/n; SE = sqrt(phat*(1-phat)/n)
p_cent = phat + crit^2/(2*n)
margin = crit*sqrt(SE^2+crit^2/(4*n^2))
u = mult*(p_cent + margin)
# Bounds
c(l, u)

[1] 0.5320789 0.7798436

One-tailed example:

x = 40; n = 60; alpha = 0.05
crit = qnorm(1-alpha)
# Continuity, set c = 0 for no continuity correction
c = 0.5
# Set scale
mult = (n/(n + crit^2))
# Right tail
phat = (x-c)/n; SE = sqrt(phat*(1-phat)/n)
p_cent = phat + crit^2/(2*n)
margin = crit*sqrt(SE^2+crit^2/(4*n^2))
l = mult*(p_cent - margin)
c(l, 1)
# Left tail
phat = (x+c)/n; SE = sqrt(phat*(1-phat)/n)
p_cent = phat + crit^2/(2*n)
margin = crit*sqrt(SE^2+crit^2/(4*n^2))
u = mult*(p_cent + margin)
c(0, u)

To get the simpler Wald confidence interval for \(p\):

For \(\hat p = \frac{x}{n}\), and standard error estimate, \(\widehat{\text{SE}} = \sqrt {\frac {\hat p(1- \hat p)}{n}}\).

For two-tailed:

\[CI = \left[\hat p - z_{\alpha/2}* \widehat{\text{SE}} \;,\; \hat p + z_{\alpha/2}* \widehat{\text{SE}} \right].\]

For right one-tailed:

\[CI = \left[\hat p - z_{\alpha}* \widehat{\text{SE}} \;,\; 1 \right].\]

For left one-tailed:

\[CI = \left[0 \;,\; \hat p + z_{\alpha}* \widehat{\text{SE}} \right].\]

x = 40; n = 60; alpha = 0.05; phat = x/n
SE = sqrt(phat*(1-phat)/n)
l = phat - qnorm(1-alpha/2)*SE
u = phat + qnorm(1-alpha/2)*SE
c(l, u)

[1] 0.5473871 0.7859463

One-tailed example:

#Right tailed
l = phat - qnorm(1-alpha)*SE
c(l, 1)
#Left tailed
u = phat + qnorm(1-alpha)*SE
c(0, u)