Please help me with this homework question. Answer all parts 6.2.19-1Question HelpIn a random sample of eleven cell phones, the mean full retail price was $510.00 and the standard deviation was $199.00. Assume the population is normally distributedand use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean µ. Interpret the results.Identify the margin of error.(Roue as needed.)dollarssquare dollarscell phones

Name: Please help me with this homework question. Answer all parts 6.2.19-1Q
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Description: Please help me with this homework question. Answer all parts 6.2.19-1Question HelpIn a random sample of eleven cell phones, the mean full retail price was $510.00 and the standard deviation was $199.00. Assume the population is normally distributedan

Given: Let n represents the sample of cell phones; that is; n=11. Let X¯ represents the sample mean full retail price; that is; X¯= $510.00. Let s represents the standard deviation; that is; s=$199.00. Here, the given level of confidence is 0.99. which implies: Level of significance=(1-0.99) =0.01 From the t table, the critical value (t*)obtained at (11-1)=10 degrees of freedom and at 0.01 level of significance is 3.169. The margin of error (E) of t distribution for the population mean μ is given as: E= t*×sn =3.169×19911 =190.14 Hence, the margin of error for the population mean μ is $190.14.The 99% confidence interval for the population mean μ is given as: C.I=X¯ ± t*sn The upper limit (U.L) of the confidence interval is given as: U.L=X¯ + t*sn =X¯ +E =510+190.14 =700.14 The lower limit (L.L) of the confidence interval is given as: L.L=X¯ - t*sn =X¯ -E =510-190.14 =319.86 Hence, the 99% confidence interval for the population mean μ is (319.86,700.14).