A community college provides the starting salaries of its graduates in a particular program as a way of attracting new students. The student enrollment office has conducted a survey of its graduates and 100 of them have volunteered their starting salaries for the jobs they have accepted. Based on this data, what is the 95% confidence interval for the average starting salary of the graduates of this program? Clearly here, we don't want to just say this is what you will make. We want to give the range in our brochures saying that our graduates make something between x and y, and that would be our interval estimates of the starting salary. Here are the 100 people that have decided to share with us what they are making. First, we will start by calculating the average salary for these graduates who've shared their numbers with us. This would be equal to average and I will pick the first cell, control, shift, down, close the parenthesis, return and it turns out the average salary is $27,092. I can also calculate the standard deviation and that would be the standard deviation.S and I would pick the same cells, and that'll be $2,534 roughly. Sample size, how many people got back to us is the number of people who responded back and that's a count of these values, and it says that it has 100. Now I can calculate the standard error and standard error is the standard deviation of the sample divided by the square root of the sample size. Now that I have this and I want a 95% confidence, then the t value is t.inverse. Again, you're doing 95%, so this going to be 0.975, 0.025 on the left tail is also added and the degrees of freedom is 100 minus 1, which is 99. So this is my critical value of t. And based on this, I can find the margin of error, and margin of error is the t value multiplied by the standard error. So now I'm ready to calculate the lower and upper bond. The lower bound on my 95% confidence interval for the mean salary is going to be the mean of this sample minus margin of error and mean of this sample plus the margin of error. So, the 95% confidence interval tells us that the average or expected salary upon graduation from this department for the population of the graduates is somewhere between 26602 to 27582. It could be any value in this interval. So now, let's look back at our problem statement and see the second part. The director of admissions believes that students wouldn't choose to enroll for the program if upon graduation, they will not make at least $26,500. What is the 95% confidence interval for proportion of graduates who make at least $26,500? So if I look at my data, I should be able to tell that roughly the average is here, so 26,500 is below the lower bound of the average. So we are probably in good shape here. But we can actually find out what proportion by getting more than 26 500. So I have re copied this salary right here and I'm going to ask makes more than 26,500 question in this column. And as before, it's going to be an if statement, and the if would say if this value is greater than 26500, then return a value of 1. These people are getting more. Otherwise return a value of zero. And I can copy this, and you can pay attention to whether you've done it correctly, so if you come to data number seven, this person made less or got an offer for less money. And if I've done this correctly, I should get a value of 0 there. And sure enough I do, I get the value of 0 for this one, and I get a value of 0 for this one. Everything else gets a value of 1, because all the other values are over 26,500. So I can just put my crosshair here double click and it will fill it up for me. Now I can find out from my data so I'm going to keep here number of people who are making more than 26500. And that would be simply the sum of values of ones that I see here because they represent people who are making more than 26500 and this would be now 61. So, 61 out of 100. So again, I knew that sample size here is 100, I already calculated that before. So then the proportion making more than 26,500 would be simply 61 out of 100, or 61%. So for 95% confidence interval, remember for proportion I would always use a z value. The z value would be based on norm.S.inverse and 0.975. And that's 1.96. And my standard error is simply square root of p times 1 minus p divided by the sample size of 100. So that's my standard error. Now that I have my standard error, I can find out what my margin of error is. And margin of error is the z value times standard error. And now, the confidence interval. The lower percentage and upper percentage of the estimate for my population proportion is simply the proportion of this sample minus margin of error, and the proportion of this sample plus the margin of error. And I find that at about 51% to 70% of graduates for May, 26,500. Now, does this mean something that I can advertise? Because I can just focus on 70% because that's what is really really appealing to me for my advertising, because the true proportion could be 51%. And if you say to someone that half of you will make 26,500, that may not be as appealing as it would've been if you had said 70%. But in reality, any value between these two is equally likely to be the true proportion. So, you have to think hard in terms of what would be the threshold. What would I have liked to see here if I had seen 80% of the people make over 26500? Would I make that advertising? Would you be happy if somebody made that advertising to you? And they really meant only 51% of the people make that. So these are the type of things you have to think about when you read a report, when somebody is telling you something, or you are going to give a report. You can really misuse statistics or misunderstand statistics if you don't pay close attention to what is being communicated.
