Answer :
Answer:
Conclusion
The average rental rate in Hong Kong are higher than in Paris
Step-by-step explanation:
From the question we are told that
The sample size for Hong Kong is [tex]n_1 = 30[/tex]
The mean for Hong Kong is [tex]\mu_h = 1,114 \ m^2[/tex]
The standard deviation for Hong Kong is[tex]s = 230[/tex]
The sample size for Paris is [tex]n_2 = 40[/tex]
The mean for Paris is [tex]\mu_p = 989 \ m^2[/tex]
The standard deviation for Paris is [tex]s_p = 195[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The Null Hypothesis is
[tex]H_o : \mu_1 - \mu_2 = 0[/tex]
The alternative hypothesis is
[tex]H_a: \mu_1 - \mu_2 >0[/tex]
The test statistics is mathematically represented as
[tex]t = \frac{\mu _1 - \mu_2 - d }{\sqrt{(\frac{s^2}{n1} )+(\frac{(s_p^2}{n_2} )}}[/tex]
Substituting values
[tex]t = \frac{1114 - 989 - 0 }{\sqrt{(\frac{230^2}{30} )+(\frac{(195^2}{40} )}}[/tex]
[tex]t = 2.399[/tex]
Since the value of the test statistics is higher than the significance level then there is enough evidence to conclude that the average rental rate in Hong Kong are higher than in Paris
Using the t-distribution, it is found that since the test statistic is more than the critical value, there is enough evidence to conclude that the mean annual lease rate is higher in Hong Kong than in Paris.
At the null hypothesis, it is tested if the average is not higher in Hong Kong, that is, the result of the subtraction if not greater than 0.
[tex]H_0: \mu_H - \mu_P \leq 0[/tex]
At the alternative hypothesis, it is tested if it is higher, that is:
[tex]H_1: \mu_H - \mu_P > 0[/tex]
The standard errors for both samples are:
[tex]s_H = \frac{230}{\sqrt{30}} = 42[/tex]
[tex]s_P = \frac{195}{\sqrt{40}} = 30.83[/tex]
The distribution of the differences has mean and standard deviation given by:
[tex]\overline{X} = \mu_H - \mu_P = 1114 - 989 = 125[/tex]
[tex]s = \sqrt{s_H^2 + s_P^2} = \sqrt{42^2 + 30.83^2} = 52.1[/tex]
The test statistic is:
[tex]t = \frac{\overline{X} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence, the value is:
[tex]t = \frac{125 - 0}{52.1}[/tex]
[tex]t = 2.4[/tex]
The critical value, for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 30 + 40 - 2 = 68 df, is [tex]t^{\ast} = 1.67[/tex]
Since the test statistic is more than the critical value, there is enough evidence to conclude that the mean annual lease rate is higher in Hong Kong than in Paris.
A similar problem is given at https://brainly.com/question/16402495