On friday evenings after 600pm it can be very difficult to g

On friday evenings after 6:00pm it can be very difficult to get a table. Suppose that the waiting time for a table on Friday evenings after 6PM is approximately normal. A random sample of 12 groups of people who dined on a Fridy evening had a mean wait time of 76.8 minutes and a standard deviation of 13.7 minutes.

1) Find and interpret a 95% CONFIDENCE INTERVAL for the mean waiting time.

2) Find and interpret a 95% PREDICTION INTERVAL for the waiting time for a future

Solution

Ans:

Let X denote the waiting time for a table.

Here, X ~ N(76.8, 13.7)

Given sample size= 12 , mean (X\') =76.8 and standard deviation (s) = 13.7

a) The 95% confidence interval for the mean waiting time is as below:

Since sample size is less than 30 and the population standard deviation is not know, we will use the t distribution instaed of the Z distribution.

(X\' - t s/sqrt(n) , X\' + t s/ sqrt(n))

= (76.8- 2.201 *13.7 / sqrt(12) , 76.8 + 2.201 *13.7 / sqrt(12))

= (68.095, 85.504)

b) 95% prediction interval for the waiting time can be found as:

The 95% prediction interval for the forecasted value Y₀^ for given x₀ is given by:

( Y₀^ - t _crit . se, Y₀^ + t _crit ?. se)

where se = s . sqrt[ 1+ 1/n + (x_{0 -} x\')^2 / SS_x )

We should be provided with the dataset from which we can calculate the estimated value of Y which is Y₀^.

Here SS_x = sum of squares for explanatory variable X.