Question

A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.05. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data.New York CityPhoenix9894.295.57292.286.8102120.185.4114.48093.785.489.775.4104.779.576.682.4106.864.398.665.591.587.78210497.774.364.959.58282.872115.2The Hypotheses for this problem are:H0: μ1 = μ2H1: μ1 ≠ μ2a) Find the p-value. Round answer to 4 decimal places.b) Choose the correct decision and summary based on the above p-value.Reject H0. There is no evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.Do not reject H0. There is evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.Do not reject H0. There is no evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.Reject H0. There is evidence that the mean temperature in New York City is different from the mean temperature in Phoenix.

Answer 1

a)
`t=\frac{83.17-92.38}{\sqrt{(12.904^2)/(17)+(15.90^2)/(18)}}=-1.886`

`df=n_(1)+n_(2)-2=17+18-2=33`

Since is a two tailed test, the p value would be:

`p_v =2*P(t_((33))<-1.886)=0.0681`

If we compare the p value and the significance level given
`\alpha=0.05` we have that
`p_v>\alpha` 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.

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

`\bar X_(1)` represent the mean for the sample New york

`\bar X_(2)` represent the mean for the sample Phoenix

`s_(1)` represent the sample standard deviation for New York

`s_(2)` represent the sample standard deviation for Phoenix

`n_(1)=17` sample size for the group New York

`n_(2)=18` 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:
`\mu_(1)=\mu_(2)`

H1:
`\mu_(1) \\eq \mu_(2)`

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:

`t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (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:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X)^2)/(n-1)}` (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:

`\bar X_(1)=83.17`
`s_(1)=12.904`

`\bar X_(2)=92.38`
`s_(2)=15.90`

And with this we can replace in formula (1) like this:

`t=\frac{83.17-92.38}{\sqrt{(12.904^2)/(17)+(15.90^2)/(18)}}=-1.886`

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:

`df=n_(1)+n_(2)-2=17+18-2=33`

Since is a two tailed test, the p value would be:

`p_v =2*P(t_((33))<-1.886)=0.0681`

If we compare the p value and the significance level given
`\alpha=0.05` we have that
`p_v>\alpha` 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.