A biologist argues that a certain species of fish found in Lake Tanzania in Africa should be classified in a different genus than the currently accepted genus for the species. A simple random sample of standard lengths of fully grown fish from the species in Lake Tanzania is given below:{116.2, 100.3, 118, 152.8, 141.3, 132.8, 95.6, 111.5, 95.2, 95.7}The population of all such lengths is assumed to be close to normally distributed. Calculate the following interval estimates using this sample (note that only one of the methods below can be used for a single sample in practice). Round answers to one decimal place accuracy.A 99% confidence interval for the mean of all fully grown fish of this type is:[type your answer...] [type your answer...]A 99% prediction interval for the length of one fully grown fish of this type is:[type your answer...] [type your answer...]

Followings are the Explanation of the question Use the Excel and find the Sample Mean and Sample Standard deviation for the given data Formula :- Using Excel 1) Sample Mean =Average() 2) Sample Standard deviation =STDEV.S()a) Use the one sample t test confidence interval for the given data, because the Population Standard unknown The provided sample mean is X= 29.2 Sample standard deviation is S = 5.9 The size of the sample is n = 33 The required confidence level is 95% Find the Level of significance α 100(1-alpha) %= 99% 100(1-alpha) = 100 (1-alpha) = 99/100 (1-alpha) = 0.99 alpha = 1-0.99 alpha = 0.01 Critical Value:- Calculate the Critical value using t table with Corresponding Alpha and Degrees of Freedom (DF) The significance level is α =0.01, and Degrees of Freedom n-1) = 10-1=9 Use the Excel and use the below formula to calculate the critical value α = 0.01 and DF= 9 Formula :- =T.INV(α,DF) = T.INV(0.01 ,9) tC=3.2498 (Note:- Put the location value of α and DF from the excel) The 99% confidence for the population means μ is computed using the following expression, CI=[X¯-tc*s/√n],[X¯+tc*s/√n]CI=[115.94-3.2498*20.57/√10],[115.94+3.2498*20.57/√10]CI=[94.8,137.1] Therefore, based on the information provided, the 99 % confidence for the population mean μ is [ 94.8< µ < 137.1]. b) Use the one sample t test Prediction interval for the given data, because the Population Standard unknown The provided sample mean is X= 29.2 Sample standard deviation is S = 5.9 The size of the sample is n = 33 The required confidence level is 95% Find the Level of significance α 100(1-alpha) %= 99% 100(1-alpha) = 100 (1-alpha) = 99/100 (1-alpha) = 0.99 alpha = 1-0.99 alpha = 0.01 Critical Value:- Calculate the Critical value using t table with Corresponding Alpha and Degrees of Freedom (DF) The significance level is α =0.01, and Degrees of Freedom n-1) = 10-1=9 Use the Excel and use the below formula to calculate the critical value α = 0.01 and DF= 9 Formula :- =T.INV(α,DF) = T.INV(0.01 ,9) tC=3.2498 (Note:- Put the location value of α and DF from the excel) The 99% prediction interval for the length of one fully grown is computed using the following expression, CI=[X¯-tc*s*√(1+1/n)],[X¯+tc*s*√(1+1/n)]CI=[115.94-3.2498*20.57*√(1+1/10)],[115.94+3.2498*20.57*√(1+1/10)]CI=[45.8,186.0] Therefore, based on the information provided, the 99 % prediction interval for the length of one fully grown is [ 45.8< µ < 186.0].