A recently published article included data from a survey of 2186 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2186 people who participated in the survey, 1312 indicated that they use social networking site to research job applicants. Based on the survey data, is there convincing evidence that fewer than two-thirds of hiring managers and HR professionals use social networking sites in this way? Use a significance level of = 0.01 to make your conclusion.

Calculate the test statistic.

Step-by-step explanation:

two-thirds = 2/3 = 0.67

We would set up the hypothesis test.

For the null hypothesis,

p = 0.67

For the alternative hypothesis,

p < 0.67

This is a left tailed test.

Considering the population proportion, probability of success, p = 0.67

q = probability of failure = 1 - p

q = 1 - 0.67 = 0.33

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 1312

n = number of samples = 2186

P = 1312/2186 = 0.6

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.6 - 0.67)/√(0.67 × 0.33)/2186 = - 6.96

From the normal distribution table, the probability value corresponding to the z score is < 0.00001

Since alpha, 0.01 > than the p value, then we would reject the null hypothesis. Therefore, at 1% significance level, there is convincing evidence that fewer than two-thirds of hiring managers and HR professionals use social networking sites in this way.

all it means is function 3