Q:

# A survey of 1010 college seniors working towards an undergraduate degree was conducted. Each student was​ asked, "Are you planning or not planning to pursue a graduate​ degree?" Of the 1010​ surveyed, 658 stated that they were planning to pursue a graduate degree. Construct and interpret a​ 98% confidence interval for the proportion of college seniors who are planning to pursue a graduate degree. Round to the nearest thousandth.

Accepted Solution

A:
Answer:The 98% confidence interval would be given (0.616;0.686).  We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686). Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  The margin of error is the range of values below and above the sample statistic in a confidence interval.  Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  Description in words of the parameter p$$p$$ represent the real population proportion of people that they were planning to pursue a graduate degree$$\hat p$$ represent the estimated proportion of people that they were planning to pursue a graduate degreen=1010 is the sample size required  $$z_{\alpha/2}$$ represent the critical value for the margin of error  The population proportion have the following distribution  $$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$$Numerical estimate for pIn order to estimate a proportion we use this formula:$$\hat p =\frac{X}{n}$$ where X represent the number of people with a characteristic and n the total sample size selected.$$\hat p=\frac{658}{1010}=0.651$$ represent the estimated proportion of people that they were planning to pursue a graduate degreeConfidence intervalThe confidence interval for a proportion is given by this formula  $$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$$  For the 98% confidence interval the value of $$\alpha=1-0.98=0.02$$ and $$\alpha/2=0.01$$, with that value we can find the quantile required for the interval in the normal standard distribution.  $$z_{\alpha/2}=2.33$$  And replacing into the confidence interval formula we got:  $$0.651 - 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.616$$  $$0.651 + 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.686$$  And the 98% confidence interval would be given (0.616;0.686).  We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).