Answer:
There is no difference between two neighboring communities in terms of their usage of social media.
Step-by-step explanation:
In this case we need to determine whether there is any difference between two neighboring communities in terms of their usage of social media.
The hypothesis can be defined as follows:
<em>H₀</em>: There is no difference between two neighboring communities in terms of their usage of social media, i.e. <em>p</em>₁ - <em>p</em>₂ = 0.
<em>Hₐ</em>: There is a significant difference between two neighboring communities in terms of their usage of social media, i.e. <em>p</em>₁ - <em>p</em>₂ ≠ 0.
The test statistic is defined as follows:
![z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%20p_%7B1%7D-%5Chat%20p_%7B2%7D%7D%7B%5Csqrt%7B%5Chat%20P%281-%5Chat%20P%29%5B%5Cfrac%7B1%7D%7Bn_%7B1%7D%7D%2B%5Cfrac%7B1%7D%7Bn_%7B2%7D%7D%5D%7D%7D)
The information provided is:
n₁ = 150
n₂ = 200
= 0.55
![\hat p_{2}=\frac{120}{200}=0.60](https://tex.z-dn.net/?f=%5Chat%20p_%7B2%7D%3D%5Cfrac%7B120%7D%7B200%7D%3D0.60)
Compute the total proportions as follows:
![\hat P=\frac{n_{1}\hat p_{1}+n_{2}\hat p_{2}}{n_{1}+n_{2}}=\frac{(150\times 0.55)+(200\times 0.60)}{150+200}=0.579](https://tex.z-dn.net/?f=%5Chat%20P%3D%5Cfrac%7Bn_%7B1%7D%5Chat%20p_%7B1%7D%2Bn_%7B2%7D%5Chat%20p_%7B2%7D%7D%7Bn_%7B1%7D%2Bn_%7B2%7D%7D%3D%5Cfrac%7B%28150%5Ctimes%200.55%29%2B%28200%5Ctimes%200.60%29%7D%7B150%2B200%7D%3D0.579)
Compute the test statistic value as follows:
![z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%20p_%7B1%7D-%5Chat%20p_%7B2%7D%7D%7B%5Csqrt%7B%5Chat%20P%281-%5Chat%20P%29%5B%5Cfrac%7B1%7D%7Bn_%7B1%7D%7D%2B%5Cfrac%7B1%7D%7Bn_%7B2%7D%7D%5D%7D%7D)
![=\frac{0.55-0.60}{\sqrt{0.579(1-0.579)[\frac{1}{150}+\frac{1}{200}]}}\\\\=-0.94](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.55-0.60%7D%7B%5Csqrt%7B0.579%281-0.579%29%5B%5Cfrac%7B1%7D%7B150%7D%2B%5Cfrac%7B1%7D%7B200%7D%5D%7D%7D%5C%5C%5C%5C%3D-0.94)
The test statistic value is -0.94.
The decision rule is:
The null hypothesis will be rejected if the <em>p</em>-value of the test is less than the significance level, <em>α</em> = 0.01.
Compute the <em>p</em>-value as follows:
![p-value=2\times P(Z](https://tex.z-dn.net/?f=p-value%3D2%5Ctimes%20P%28Z%3C-0.94%29%5C%5C%3D2%5Ctimes%200.17361%5C%5C%3D0.34722%5C%5C%5Capprox%200.35)
*Use a <em>z</em>-table.
The <em>p</em>-value of the test is 0.35.
<em>p</em>-value = 0.35 > <em>α</em> = 0.01.
The null hypothesis will not be rejected at 1% significance level.
Thus, it can be concluded that there is no difference between two neighboring communities in terms of their usage of social media.