Question

auto insurance: claims anystate auto insurance company took a random sample of 370 insurance claims paid out during a 1 - year period. the average claim paid was $1570. assume $sigma=$ 250$. find 0.90 and 0.99 confidence intervals for the mean claim payment.

psychology: closure three experiments investigating the relationship between need for cognitive closure and persuasion were reported in \motivated resistance and openness to persuasion in the presence or absence of prior information\ by a. w. kruglanski (journal of personality and social psychology, vol. 65, no. 5, pp. 861 - 874). part of the study involved administering a
eed for closure scale\ to a group of students enrolled in an introductory psychology course. the
eed for closure scale\ has scores ranging from 101 to 201. for the 73 students in the highest quartile of the distribution, the mean score was $\bar{x}=178.70$. assume a population standard deviation of $sigma = 7.81$. these students were all classified as high on their need for closure. assume that the 73 students represent a random sample of all students who are classified as high on their need for closure. find a 95% confidence interval for the population mean score $mu$ on the
eed for closure scale\ for all students with a high need for closure.

Explanation:

Step1: Determine the formula for confidence interval

The formula for a confidence interval for the population mean when the population standard deviation \(\sigma\) is known is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)

Step2: Calculate the 0.90 confidence interval for the auto - insurance problem

• For a 0.90 confidence interval, \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2 = 0.05\). The \(z\) - value \(z_{\alpha/2}=z_{0.05}\). From the standard normal table, \(z_{0.05} = 1.645\)
• Given \(\bar{x}=1570\), \(\sigma = 250\), \(n = 370\)
• The margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.645\times\frac{250}{\sqrt{370}}\approx1.645\times\frac{250}{19.235}\approx1.645\times12.997\approx21.39\)
• The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\), so \(1570 - 21.39<\mu<1570+21.39\), which is \(1548.61<\mu<1591.39\)

Step3: Calculate the 0.99 confidence interval for the auto - insurance problem

• For a 0.99 confidence interval, \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\). The \(z\) - value \(z_{\alpha/2}=z_{0.005}\). From the standard normal table, \(z_{0.005}=2.576\)
• The margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{250}{\sqrt{370}}\approx2.576\times\frac{250}{19.235}\approx2.576\times12.997\approx33.57\)
• The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), so \(1570-33.57<\mu<1570 + 33.57\), which is \(1536.43<\mu<1603.57\)

Step4: Calculate the 0.95 confidence interval for the psychology problem

• For a 0.95 confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2 = 0.025\). The \(z\) - value \(z_{\alpha/2}=z_{0.025}\). From the standard normal table, \(z_{0.025}=1.96\)
• Given \(\bar{x}=178.70\), \(\sigma = 7.81\), \(n = 73\)
• The margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{7.81}{\sqrt{73}}\approx1.96\times\frac{7.81}{8.544}\approx1.96\times0.914\approx1.8\)
• The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), so \(178.70-1.8<\mu<178.70 + 1.8\), which is \(176.9<\mu<180.5\)

Answer:

• For the auto - insurance 0.90 confidence interval: \((1548.61,1591.39)\)
• For the auto - insurance 0.99 confidence interval: \((1536.43,1603.57)\)
• For the psychology 0.95 confidence interval: \((176.9,180.5)\)