A team of software engineers are testing the time taken for a particular type of modern computer to execute a complicated algorithm for factoring large numbers. They would like to estimate the mean time taken for a computer to execute the algorithm. A random sample of 61 times are collected. The mean time in this sample is 804.9 seconds and the sample standard deviation is found to be 85.1. Calculate the 99% confidence interval for the mean time taken to execute the algorithm. Give your answers to 2 decimal places.

99% confidence interval for the mean time taken to execute the algorithm is (775.92 , 833.88).

Step-by-step explanation:

We are given that a random sample of 61 times are collected. The mean time in this sample is 804.9 seconds and the sample standard deviation is found to be 85.1.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 99% confidence interval for the population variance is given by;

       P.Q. = ~

where, = sample mean time = 804.9 seconds

            = population mean time

            s = sample standard deviation = 85.1 seconds

            n = sample size = 61

So, 99% confidence interval for population mean, is;

P(-2.66 < < 2.66) = 0.99 {As the t table at 60 degree of freedom

                                              gives critical values of -2.66 & 2.66}

P(-2.66 < < 2.66) = 0.99

P( < < ) = 0.99

P( < < ) = 0.99

99% confidence interval for = ( , )

                                                 = ( , )

                                                 = (775.92 , 833.88)

Therefore, 99% confidence interval for the mean time taken to execute the algorithm is (775.92 , 833.88).

where is the rest of the

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answer

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