# URGENT!!! Statistic question

In 2010 London introduced the Barclays London Bicycle Hire scheme that gives people the opportunity to hire bicycles from various docking stations in London. The scheme intends to promote the use of bicycles in London as a way to reduce traffic, congestions and to improve people’s fitness. The scheme is designed to favour the use of the bicycles for a short period of time, usually less than 60 minutes. In fact, the first 30 minutes are free and it costs £1 to hire the bike for the first hour. The cost sharply increases to £4 for the first hour and a half and £6 for up to two hours. More information about the scheme can be found from the following website:

You are asked to use inferential statistics to estimate on the provided sample the average length of bicycle hire. To this end, on Blackboard you will be able to access the excel file which contains a large amount of data about the actual usage of the bicycles.

(Because it is in excel, so I just extract some useful data given in the excel)

number of sample=51347

sample mean=21.16

sample SD=151.13

a) Suppose you have completed your data collection and you now have available the dataset “EC1012 Cycle Hire Data”. Suggest an unbiased estimator and use it to provide an estimate of the average length of the bike hire. (25 marks)

b) Construct a confidence interval that has a high probability of capturing the average length of the bike hire. (25 marks)

c) Set up a hypothesis test to test whether the average length of the bike hire is consistent with Transport for London expectation that bikes should be hired for no longer than 60 minutes. (25 marks)

Thank you so mcuh

should we use sample SD when calculating Z-value?

should it be Z=(21.16-60)/(151.13/(51347}^1/2)? but I got 58.01, is it possible to have a Z-value large like this?

### 1 個解答

• 最佳解答

I think the sample mean is in term of minutes.

Let X_i be the ith individual data.

The unbiased estimator should be /bar X_i/51347

an estimate of the average length of the bike hire=21.16

b) As you did not specify whether it is a 95% or 90% confidence interval,I guess it is 95% confidence interval.

21.16+-1.96X151.13/(51347)^(1/2)

If it is 90% interval,replace 1.96 with 1.645

(c)H_0:the average length of the bike hire =60

H_1:the average length of the bike hire >60

The test statistic:

Z=(21.16-60)/((51347)^(1/2))

I do not know what your alpha is .

2014-03-21 07:36:08 補充：

We only look at the data.

We have found that the difference between 60 and the sample mean is great.

Hence,it is possible

2014-03-21 07:36:57 補充：

PS: I was sorry that there was a typo in my answer.