Answer :
Final answer:
To assess whether the mean temperatures in New York City and Phoenix are different, a two-sample t-test is used. The p-value results from this test dictate whether to reject the null hypothesis or not, based on whether it is less than the significance level of 0.05.
Explanation:
To test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix, we need to conduct a two-sample t-test with the null hypothesis H0: μ1 = μ2 and the alternative hypothesis H1: μ1 ≠ μ2, where μ1 is the mean temperature in New York City and μ2 is the mean temperature in Phoenix. Since the samples have unequal variances and are assumed to be normally distributed, we use the Welch's t-test, which does not assume equal population variances.
First, we calculate the sample means, variances, and the Welch's t-statistic. Then, we find the degrees of freedom using the approximation method specific for the Welch's t-test. With the calculated t-statistic and degrees of freedom, we look up the critical value or use a statistical software to obtain the corresponding p-value.
If the p-value is less than the significance level α of 0.05, we reject the null hypothesis and conclude there is evidence that the mean temperatures are different. If the p-value is greater than 0.05, we do not reject the null hypothesis and conclude there is no evidence of a difference in mean temperatures between the two cities.
Answer:
a) [tex]t=\frac{83.17-92.38}{\sqrt{\frac{12.904^2}{17}+\frac{15.90^2}{18}}}=-1.886[/tex]
[tex]df=n_{1}+n_{2}-2=17+18-2=33[/tex]
Since is a two tailed test, the p value would be:
[tex]p_v =2*P(t_{(33)}<-1.886)=0.0681[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we have that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Do not reject H0. There is no evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.
Step-by-step explanation:
Data given and notation
For this case we have the following Data:
New york: 8,95.5,92.2,102,85.4,80,85.4,75.4,79.5,82.4,64.3,65.5,87.7,104,74.3,59.5,82.8
Phoenix:94.2,72,86.8,120.1,114.4,93.7,89.7,104.7,76.6,106.8,98.6,91.5,82,97.7,64.9,82,72,115.2
[tex]\bar X_{1}[/tex] represent the mean for the sample New york
[tex]\bar X_{2}[/tex] represent the mean for the sample Phoenix
[tex]s_{1}[/tex] represent the sample standard deviation for New York
[tex]s_{2}[/tex] represent the sample standard deviation for Phoenix
[tex]n_{1}=17[/tex] sample size for the group New York
[tex]n_{2}=18[/tex] sample size for the group Phoneix
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
H0:[tex]\mu_{1}=\mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n-1}}[/tex] (3)
Calculate the statistic
First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:
[tex]\bar X_{1}=83.17[/tex] [tex]s_{1}=12.904[/tex]
[tex]\bar X_{2}=92.38[/tex] [tex]s_{2}=15.90[/tex]
And with this we can replace in formula (1) like this:
[tex]t=\frac{83.17-92.38}{\sqrt{\frac{12.904^2}{17}+\frac{15.90^2}{18}}}=-1.886[/tex]
Statistical decision
For this case we have a significance level of , now we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{1}+n_{2}-2=17+18-2=33[/tex]
Since is a two tailed test, the p value would be:
[tex]p_v =2*P(t_{(33)}<-1.886)=0.0681[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we have that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
And the best conclusion for this case would be:
Do not reject H0. There is no evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.