# Thread: Hypothesis testing

1. ## Hypothesis testing

Determine whether CCResort has been successful in meeting their key performance indicators stated in the business plan as:
1- More than 40% of their customers stay for a full week (ie. 7 days)
2- The average customer spends more than \$ 255 per day in excess of accomodation

(the full data sheet has been attached)

How would i formulate a hypothesis test, confidence interval for the following, considering that
-102 customers of the surveyed 200 stayed for 1 week

 Mean Expenditure 1170.7 Median Expenditure 1254 Mode Expenditure 428 Variance 468871

2. ## Re: Hypothesis testing

Hey tjang.

Hint: If you are getting the distribution of the mean then the variance of the estimator is Sample_Variance/#Number of observations in sample.

In other words (x_bar - mu)/[Sample_Variance/Number_Of_observations] ~ t_(n-1) = t distribution with (n-1) degrees of freedom.

You can then re-arrange to get the distribution of mu and test the hypothesis mu > 255 if your variables correspond to daily spending.

3. ## Re: Hypothesis testing

thanks chiro, i'll still a bit confused with what your saying
do you mind showing some working out
thanks in advance! =D

4. ## Re: Hypothesis testing

In your example:

Variance = 468871.2
Number of Observations = 102 = n
n - 1 = 101
x_bar = 1170.705
SE = SQRT(468871.2/102)
Since mean > 255 you get

Re-arranging you get a confidence interval of ((1170.705 - 255) - 1.96*SE, infinity) and the probability for this region is your p-value.

If