Question

A firm wanted to see if there was a significant difference between two neighboring communities in terms of their usage of social media. A random survey of 150 households in the first community showed the 55% of the respondents used social media frequently. In the second community, of a random survey of 200 households, 120 used social media frequently. Assuming a significance level of .01, test the hypothesis that the two communities are statistically equivalent with respect to the usage of social media.

Answer 1

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:  

H₀: There is no difference between two neighboring communities in terms of their usage of social media, i.e. p₁ - p₂ = 0.  

Hₐ: There is a significant difference between two neighboring communities in terms of their usage of social media, i.e. p₁ - p₂ ≠ 0.  

The test statistic is defined as follows:

 $z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

The information provided is:

n₁ = 150

n₂ = 200

$\hat p_{1}$ = 0.55

$\hat p_{2}=\frac{120}{200}=0.60$

Compute the total proportions as follows:

 $\hat P=\frac{n_{1}\hat p_{1}+n_{2}\hat p_{2}}{n_{1}+n_{2}}=\frac{(150\times 0.55)+(200\times 0.60)}{150+200}=0.579$

Compute the test statistic value as follows:

 $z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

    $=\frac{0.55-0.60}{\sqrt{0.579(1-0.579)[\frac{1}{150}+\frac{1}{200}]}}\\\\=-0.94$

The test statistic value is -0.94.

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level, α = 0.01.  

Compute the p-value as follows:

 $p-value=2\times P(Z$

*Use a z-table.

The p-value of the test is 0.35.

p-value = 0.35 > α = 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.