93.4k views
4 votes
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

1 Answer

3 votes

Answer:


t=\frac{(1860-1935)-0}{\sqrt{(126^2)/(36)+(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.

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)\\eq 0

The statistic is given by:


t=\frac{(\bar X_(1)-\bar X_(2))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\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{(126^2)/(36)+(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.

User NiziL
by
3.5k points