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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  1170.705 
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_(n1) = t distribution with (n1) degrees of freedom.
You can then rearrange to get the distribution of mu and test the hypothesis mu > 255 if your variables correspond to daily spending.

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

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
Rearranging you get a confidence interval of ((1170.705  255)  1.96*SE, infinity) and the probability for this region is your pvalue.
If