Question

A research group is curious about their city’s participation in flu shot clinics. They suspect that there is a difference between average amount of attendees at their city’s flu shot clinics when compared to their sister city. However, they must conduct a study to determine if that is true. From a sample of 20 clinics in the sister city, the sample mean of the clinic attendees is 156 and the sample standard deviation is 36. From a sample of 22 clinics in the researchers’ city, the sample mean of clinic attendees is 162 and the sample standard deviation is 41.

Determine if the researchers should accept or reject H0 under alpha = 0.05.

Note that the standard error is 11.8831 and the graph is two-tailed.

Answer 1

`t=\frac{(162-156)-0}{\sqrt{(41^2)/(22)+(36^2)/(20)}}}=0.505`

`p_v =2*P(t_(40)>0.505)=0.61633`

Comparing the p value with the significance level
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, then the difference between the two groups is NOT significantly different.

Explanation:

Data given and notation

`\bar X_(1)=162` represent the mean for sample of clinic attendees

`\bar X_(2)=156` represent the mean for the sister city

`s_(1)=41` represent the sample standard deviation for 1

`s_(2)=36` represent the sample standard deviation for 2

`n_(1)=22` sample size for the group 2

`n_(2)=20` sample size for the group 2

`\alpha=0.05` Significance level provided

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 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`

We don't have the population standard deviation's, 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))-\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=22+20-2=40`

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.

With the info given we can replace in formula (1) like this:

`t=\frac{(162-156)-0}{\sqrt{(41^2)/(22)+(36^2)/(20)}}}=0.505`

P value

Since is a bilateral test the p value would be:

`p_v =2*P(t_(40)>0.505)=0.61633`

Comparing the p value with the significance level
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, then the difference between the two groups is NOT significantly different.