Question

An ambulance agency claims that the mean length of service times is 10 minutes with SD 3 minutes. An investigator suspects that the claim is wrong. She takes a random sample of 20 services and finds the mean length of service times is 12 minutes with a SD 5 minutes. Assume that the service time of the ambulance follows normal distribution.
1) What is the probability that the mean length of service times is less than or equal to 12 minutes
2) Find 99% confidence interval for standard deviation of the mean length of service times.

Answer 1

(1) Probability that the mean length of service times is less than or equal to 12 minutes is 0.9986.

(2) 99% confidence interval for standard deviation of the mean length of service times is [3.51 minutes , 8.33 minutes].

Step-by-step explanation:

We are given that an ambulance agency claims that the mean length of service times is 10 minutes with SD 3 minutes.

An investigator takes a random sample of 20 services and finds the mean length of service times is 12 minutes with a SD 5 minutes.

Let $\bar X$ = sample mean length of service times

The z score probability distribution for sample mean is given by;

                            Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean length of service times = 10 min

            $\sigma$ = population standard deviation = 3 minutes

            n = sample of services = 20

(1) Probability that the mean length of service times is less than or equal to 12 minutes is given by = P($\bar X$ $\leq$ 12 minutes)

      P($\bar X$ $\leq$ 12 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{12-10}{\frac{3}{\sqrt{20} } }$ ) = P(Z $\leq$ 2.98) = 0.9986

The above probability is calculated by looking at the value of x = 2.98 in the z table which has an area of 0.9986.

(2) The Pivotal quantity for 95% confidence interval for the population standard deviation is given by;

                                   P.Q. =  $\frac{(n-1)s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

where, $s^{2}$ = sample variance = $5^{2}$ = 25 minutes

            n = sample of services = 20

            $\sigma$ = population standard deviation

Here for constructing 99% confidence interval we have used One-sample chi-square test statistics.

So, 99% confidence interval for the population standard deviation, $\sigma$ is ;

P(6.84 < $\chi^{2}__1_9$ < 38.58) = 0.99  {As the critical value of chi at 19 degree

                                                  of freedom are 6.84 & 38.58}  

P(6.84 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 38.58) = 0.99

P( $\frac{ 6.84}{(n-1)s^{2}}$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 38.58}{(n-1)s^{2}}$ ) = 0.99

P( $\frac{(n-1)s^{2}}{38.58}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2}}{6.84}$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2}}{38.58}$ , $\frac{(n-1)s^{2}}{6.84}$ ]

                                                   = [ $\frac{(20-1)\times 25}{38.58}$ , $\frac{(20-1)\times 25}{6.84}$ ]

                                                   = [12.31 , 69.44]

99% confidence interval for $\sigma$  = [ $\sqrt{12.31}$ , $\sqrt{69.44}$ ]

                                                   = [3.51 , 8.33]

Therefore, 99% confidence interval for standard deviation of the mean length of service times is [3.51 minutes , 8.33 minutes].