Good Example Of Statistical Research Report

a)
In this research I will use the basics of statistics and probability theory to test two hypotheses. There will be one test for proportion and one test for comparing means. All the calculations will be completed in Minitab 16.
b)
1)

Descriptive Statistics: Duration

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Duration 430382 0 1309 12,7 8314 0,0 480 780 1260 1228320
Mean of “Duration” is 1309 minutes
Also we can calculate that among 430382 observations, there are 707 EndStations with the name “Queen Marys, Mile End”
Hence, the sample proportion is:
p=707430382≈0.001643=0.1643%
c)-e)

Here I use tests to check our hypothesis and also show confidence intervals.

For Duration (1 sample t-test comparing sample mean with hypothesized mean of 30):
H0: μ=30Ha: μ>30a=0.01

One-Sample T: Duration

Test of mu = 30 vs > 30

95% Lower

Variable N Mean StDev SE Mean Bound T P
Duration 430382 1309,1 8314,5 12,7 1288,2 100,92 0,000
95% Confidence interval for Duration: (1284,2; 1333,9)
This is one-tailed test. Since p-value of the test is lesser than 0.001, we can reject the null hypothesis and conclude that the average length of the bike hire is significantly longer than 30 minutes (at 1% level of significance).

For “EndStation name”:

95% confidence interval is: (0,001524; 0,001768)
The test is one-sample t-test for proportion:
H0: p=0.005Ha: p>0.005a=0.01

Test and CI for One Proportion

Test of p = 0,005 vs p > 0,005

95% Lower Exact

Sample X N Sample p Bound P-Value
1 707 430382 0,001643 0,001543 1,000
Since p-value is 1, we failed to reject the null hypothesis. We can say that there is no evidence to say that the proportion of bike rides taking place at the “Queen Marys, Mile End” docking station is significantly greater than the 0.5% (at 1% level of confidence). In fact, it is significantly lesser than 0.5%