In 2018, during tutorials, we collected the heart rates of MATH1041 students in bpm (beats per minutes). Here are the values lwe got from my students: 84, 96, 78, BB, 67, so, 90, 90, so, 73, E35, 76, 74, B4, 96, 78, BB, 67, so, 90, 90, BO, 73, 85, 76, 74 Let's use the data above to estimate ,u, the true mean heart rate of ALL MATH1041 students. Note that the true standard deviation for the heart rate ofALL MATH1041 students is not known. a) A point estimate ofp is one single number which estimates p. As a point estimate of ,u, it is better to use Q the mean of all the values in the list above 0 a randomly chosen value in the list above Enter your best guess for ,u. {that is, your point estimate of pa} : $51573 bpm {Enter your answer correct to at least three decimal places} b) Now we would like to estimate ,u using an interval rather than just one number, in other words, we want a confidence interval. i) For the sample above, the sample standard deviation is : $ bpm (Enter your answer correct to at least three decimal places) b) Now we would like to estimate p. using an interval rather than just one number, in other words, we want a confidence interval. i) For the sample above, the sample standard deviation is : 515?? bpm (Enter your answer correct to at least three decimal places) ii) We want a 95 % condence interval. To nd it, we need to find the value of the number 15* in the formula (see lecture notes). We get it using the tdistribution v with $515? degrees of freedom. 'rfP'. Help with RStudio: If you have stored the above values in an object called heartbeats, you can get the number _ll" of objects in the list using: length(heartbeats} t* = g bpm (Enter your answer correct to at least three decimal places) iii) Now we can calculate the values of the endpoints of our realised condence interval and use them to ll in the blanks: We are 95% confident that the true heartbeat of MATH1041 students is between $1751\" and git}? beats per minutes. (Enter these values correct to one decimal place)