Statistical Hypothesis Test on Guest Stay in Accommodations Data

This assignment demonstrates a two-sample hypothesis test to compare the average time spent by guests in hotels versus bed and breakfasts. The null hypothesis states that there is no difference in the average time, while the alternative hypothesis suggests a difference. The independent two-sa...Read More

The two samples hypothesis test is conducted in six steps as shown below.
Step 1: Null and alternative hypotheses
Null hypothesis H0 : μHotel=μB∧B
Alternative hypothesis H0 : μHotel ≠ μB∧B
Step 2: Value of test statistic
The independent two sample t test for unequal variance would be used and the t statistic is
computed as highlighted below.
t stat= x 1−x 2
√ s1
2
n1
+ s2
2
n2
t stat= 7.5391−7.7150
√ ( 1.3212 ) 2
32 + ( 1.5794 ) 2
16
=−0.3835
Step 3: The significance level
Assuming a significance level of 5%.
Step 4: Degree of freedom and the p value
Degree of freedom df =
( s1
2
n1
+ s2
2
n2 )
2
( s1
2
n1 )
2
n1−1 +
( s2
2
n2 )
2
n2 −1
df =
( ( 1.3212 ) 2
32 + ( 1.5794 ) 2
16 )
2
( ( 1.3212 ) 2
32 )
2
32−1 + ( ( 1.5794 ) 2
16 )
2
16−1
=25.804 26

The p value for 26 degree of freedom, -0.3835 t test and two tailed test comes out to be
0.7044.
Step 5: Result
Null hypothesis would be rejected when the p value is lower than the significance level. It
can be seen from the above analysis that p value is 0.7044 which is higher than level of
significance and hence, insufficient evidence is present to reject the null hypothesis. Hence,
alternative hypothesis would not be accepted.
Step 6: Conclusion