Respuesta :
Answer:
The 90% confidence interval for the proportion of job recruiters who reassessed a job candidate after viewing his/her social media content is (0.4137, 0.5063).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
In a survey of 314 job recruiters, 46% said that they had reassessed a job candidate after viewing his/her social media profile and content. (This includes both positive and negative reassessments of the job candidate.)
This means that [tex]n = 314, \pi = 0.46[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 1.645\sqrt{\frac{0.46*0.54}{314}} = 0.4137[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 1.645\sqrt{\frac{0.46*0.54}{314}} = 0.5063[/tex]
The 90% confidence interval for the proportion of job recruiters who reassessed a job candidate after viewing his/her social media content is (0.4137, 0.5063).
