Question

A random sample of 36 tourists in the Grand Bahamas showed that they spent an average of $1,860 (in a week) with a standard deviation of $126; and a sample of 64 tourists in New Province showed that they spent an average of $1,935 (in a week) with a standard deviation of $138. Is there any significant difference between the average expenditures of those who visited the two islands

Answer 1

Answer:

$t=\frac{(1860-1935)-0}{\sqrt{\frac{126^2}{36}+\frac{138^2}{64}}}}=-2.76$  

Now we can calculate the p value:

$p_v =2*P(t_{98}<-2.76)=0.0069$  

For this case the p value is a very lowe value so then we have enough evidence to reject the null hypothesis. Since the p value is a very low value.

Step-by-step explanation:

Information given

$\bar X_{1}=1860$ represent the mean for Grand Bahamas

$\bar X_{2}=1935$ represent the mean for New Province  

$s_{1}=126$ represent the sample standard deviation for Grand Bahamas

$s_{2}=138$ represent the sample standard deviation for New Bahamas

$n_{1}=36$ sample size for the group Grand Bahamas

$n_{2}=64$ sample size for the group New Bahamas

t would represent the statistic

System of hypothesis

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}=0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

The statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=36+64-2=98$  

The statistic can be calculated like this:

$t=\frac{(1860-1935)-0}{\sqrt{\frac{126^2}{36}+\frac{138^2}{64}}}}=-2.76$  

Now we can calculate the p value:

$p_v =2*P(t_{98}<-2.76)=0.0069$  

For this case the p value is a very lowe value so then we have enough evidence to reject the null hypothesis. Since the p value is a very low value.