Chapter 8. Testing for Differences in Mean and Proportion: Independent Proportions Z-test
Independent Proportions Z-test: Test Statistic and p-value
Independent proportions Z-test: Test Statistic
Let #X_1# denote the number of successes in the first sample and #X_2# the number of successes in the second sample. Then #\hat{p}_1# and #\hat{p}_2# are the sample proportions:
\[\hat{p}_1 = \cfrac{X_1}{n_1} \phantom{000000} \hat{p}_2 = \cfrac{X_2}{n_2}\]
Besides the individual sample proportions, we will also need the pooled sample proportion #\hat{p}# in order to calculate the test statistic:
\[\hat{p} = \cfrac{X_1+X_2}{n_1+n_2}\]
The test statistic of a independent proportions #Z#-test is denoted #Z# and is computed with the following formula:
\[Z=\cfrac{(\hat{p}_1-\hat{p}_2) - (\pi_1 - \pi_2)}{s_{(\hat{p}_1 - \hat{p}_2)}} = \cfrac{\hat{p}_1-\hat{p}_2 }{\sqrt{\hat{p}\cdot(1-\hat{p})\cdot(\cfrac{1}{n_1}+\cfrac{1}{n_2})}}\]
where #s_{(\hat{p}_1 - \hat{p}_2)}# is the standard error of the proportion difference.
When both samples are large #(n_1 \geq 30 \text{ and } n_2 \geq 30)#, the #Z#-statistic follows the Standard Normal Distribution under the null hypothesis of the test:
\[Z \sim N(0,1)\]
Calculating the p-value of an independent proportions Z-test with Statistical Software
The calculation of the #p#-value of an independent proportions #Z#-test is dependent on the direction of the test and can be performed using either Excel or R.
To calculate the #p#-value of an independent proportions #Z#-test for #\pi_1 - \pi_2# in Excel, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{000000}H_0&\phantom{000000}H_a&\phantom{0000000000}\text{Excel Command}\\
\hline
\text{Two-tailed}&H_0:\pi_1 - \pi_2 = 0&H_a:\pi_1 - \pi_2 \neq 0&=2 \text{ * }(1 \text{ - }\text{NORM.DIST}(\text{ABS}(z),0,1,1))\\
\text{Left-tailed}&H_0:\pi_1 - \pi_2 \geq 0&H_a:\pi_1 - \pi_2 \lt 0&=\text{NORM.DIST}(z,0,1,1)\\
\text{Right-tailed}&H_0:\pi_1 - \pi_2 \leq 0&H_a:\pi_1 - \pi_2 \gt 0&=1 \text{ - }\text{NORM.DIST}(z,0,1,1)\\
\end{array}\]
To calculate the #p#-value of an independent proportions #Z#-test for #\pi_1 - \pi_2# in R, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{000000}H_0&\phantom{000000}H_a&\phantom{0000000000}\text{R Command}\\
\hline
\text{Two-tailed}&H_0:\pi_1 - \pi_2 = 0&H_a:\pi_1 - \pi_2 \neq 0&2 \text{ * }\text{pnorm}(\text{abs}(z),0,1, \text{FALSE})\\
\text{Left-tailed}&H_0:\pi_1 - \pi_2 \geq 0&H_a:\pi_1 - \pi_2 \lt 0&\text{pnorm}(z,0,1, \text{TRUE})\\
\text{Right-tailed}&H_0:\pi_1 - \pi_2 \leq 0&H_a:\pi_1 - \pi_2 \gt 0&\text{pnorm}(z,0,1, \text{FALSE})\\
\end{array}\]
If #p \leq \alpha#, reject #H_0# and conclude #H_a#. Otherwise, do not reject #H_0#.
The researcher plans on using an independent proportions #Z#-test to determine whether or not there is a significant difference between the morning and evening on-time arrival rate, at the #\alpha = 0.02# level of significance.
Out of the #96# morning trains, #X_1=78# arrived on time. Out of the #102# evening trains, #X_2=92# arrived on time.
Calculate the #p#-value of the test and make a decision regarding #H_0: \pi_1 - \pi_2 = 0#. Round your answer to #3# decimal places.
#p=0.071#
On the basis of this #p#-value, #H_0# should not be rejected, because #\,p# #\gt# #\alpha#.
There are a number of different ways we can calculate the #p#-value of the test. Click on one of the panels to toggle a specific solution.
Compute the sample proportions #\hat{p}_1# and #\hat{p}_2#:
\[\hat{p}_1=\cfrac{X_1}{n_1}=\cfrac{78}{96}=0.81250\\
\hat{p}_2=\cfrac{X_2}{n_2}=\cfrac{92}{102}=0.90196\]
Compute the pooled sample proportion #\hat{p}#:
\[\hat{p}=\cfrac{X_1 + X_2 }{n_1 + n_2}=\cfrac{78 + 92}{96 + 102}=0.85859\]
Compute the #Z#-statistic:
\[z=\cfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p}) \cdot \bigg(\cfrac{1}{n_1}+\cfrac{1}{n_2} \bigg)}}
=\cfrac{0.81250 - 0.90196}{\sqrt{0.85859 \cdot (1-0.85859) \cdot \bigg(\cfrac{1}{96}+\cfrac{1}{102} \bigg)}}=-1.8055\]
Since both #n_1# and #n_2# are considered large (#\gt 30#), the Central Limit Theorem applies and we know that the test statistic
\[Z=\cfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p}) \cdot \bigg(\cfrac{1}{n_1}+\cfrac{1}{n_2} \bigg)}}\]
approximately has the Standard Normal Distribution, under the assumption that #H_0# is true.
For a two-tailed #Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in Excel, make use of the following function:
NORM.DIST(x, mean, standard_dev, cumulative)
- x: The value at which you wish to evaluate the distribution function.
- mean: The mean of the distribution.
- standard_dev: The standard deviation of the distribution.
- cumulative: A logical value that determines the form of the function.
- TRUE - uses the cumulative distribution function, #\mathbb{P}(X \leq x)#
- FALSE - uses the probability density function
Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:
\[
=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(z),0,1,1))\\
\downarrow\\
=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(\text{-}1.8055),0,1,1))
\]
This gives:
\[p = 0.071\]
Since #\,p# #\gt# #\alpha#, #H_0: \pi_1 - \pi_2 = 0# should not be rejected.
Compute the sample proportions #\hat{p}_1# and #\hat{p}_2#:
\[\hat{p}_1=\cfrac{X_1}{n_1}=\cfrac{78}{96}=0.81250\\
\hat{p}_2=\cfrac{X_2}{n_2}=\cfrac{92}{102}=0.90196\]
Compute the pooled sample proportion #\hat{p}#:
\[\hat{p}=\cfrac{X_1 + X_2 }{n_1 + n_2}=\cfrac{78 + 92}{96 + 102}=0.85859\]
Compute the #Z#-statistic:
\[z=\cfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p}) \cdot \bigg(\cfrac{1}{n_1}+\cfrac{1}{n_2} \bigg)}}
=\cfrac{0.81250 - 0.90196}{\sqrt{0.85859 \cdot (1-0.85859) \cdot \bigg(\cfrac{1}{96}+\cfrac{1}{102} \bigg)}}=-1.8055\]
Since both #n_1# and #n_2# are considered large (#\gt 30#), the Central Limit Theorem applies and we know that the test statistic
\[Z=\cfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p}) \cdot \bigg(\cfrac{1}{n_1}+\cfrac{1}{n_2} \bigg)}}\]
approximately has the Standard Normal Distribution, under the assumption that #H_0# is true.
For a two-tailed #Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in R, make use of the following function:
pnorm(q, mean, sd, lower.tail)
- q: The value at which you wish to evaluate the distribution function.
- mean: The mean of the distribution.
- sd: The standard deviation of the distribution.
- lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.
Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:
\[
2 \text{ * } \text{pnorm}(q = \text{abs}(z), mean = 0, sd = 1,lower.tail = \text{FALSE})\\
\downarrow\\
2\text{ * } \text{pnorm}(q = \text{abs}(\text{-}1.8055), mean = 0, sd = 1,lower.tail = \text{FALSE})
\]
This gives:
\[p = 0.071\]
Since #\,p# #\gt# #\alpha#, #H_0: \pi_1 - \pi_2 = 0# should not be rejected.
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