A firm wanted to see if there was a significant difference between two neighbori

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. A sample of 400 female consumers of a product had a mean age of 33 with a standard deviation of 5, and a sample of 500 male consumers had a mean age of 32 with a standard deviation of 3. Can the mean ages of the female and male populations of consumers be considered significantly different at a significance level of .05 (meaning at 95% confidence level)

Explanation / Answer

1) The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Here P1 = 0.55, P2 = 120/200 = 0.6

Here the significance level is given as 0.01

Now we need to calculate the pooled sample proportion and the standard error.

Pooled sample proportion can be calculated as below:

p = (p1 * n1 + p2 * n2) / (n1 + n2)

Here p1 = 0.55, p2 = 0.6, n1 = 150, n2 = 200

So p = (0.55 * 150 + 0.60 * 200)/ (150 + 200) = 0.5785714

Standard error can be calculated as below:

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = sqrt{ 0.5785714 * (1 - 0.5785714) * [ (1/150) + (1/200) ] }

= sqrt{ 0.2438265 * [0.01166667] }

= sqrt{ 0.002844643 } = 0.05333519

Now we can calculate the z score as below:

z = (p1 - p2) / SE

= (0.55 - 0.6)/0.05333519

= -0.9374674

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.9374674 or greater than 0.9374674.

From the z table calculator, we can see that

P(z < -0.9374674) = 0.17426, and P(z > 0.9374674) = 0.17426. Thus, the P-value = 0.17426 + 0.17426 = 0.34852

As the p value is greater than 0.01, we can not reject the null hypothesis.

That means the two communities are statistically equivalent with respect to the usage of social media.

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