Question

A journalist reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. Perform a hypothesis test to determine if a difference exists between the average time an American spends eating lunch when compared to a person from France. The following data represents the​ time, in​ minutes, that random French and American diners spent at lunch. Assume that the population variances are equal. Assume Population 1 is defined as French diners and Population 2 is defined as American diners. What is the test statistic for this hypothesis​ test?

American

21

17

17

20

25

16

20

16

French

24

18

20

28

18

29

17

Answer 1

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{S_p\sqrt{(1)/(n_1)+(1)/(n_2)}}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom and the pooled variance
`S^2_p` is given by this formula:

`\S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

`t=\frac{19 -22)-(0)}{4.095\sqrt{(1)/(8)+(1)/(7)}}=-1.416`

Explanation:

Data given

American: 21,17,17,20,25,16,20,16 (Sample 1)

French: 24,18,20,28,18,29,17 (Sample 2)

When we have two independent samples from two normal distributions with equal variances we are assuming that

`\sigma^2_1 =\sigma^2_2 =\sigma^2`

And the statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{S_p\sqrt{(1)/(n_1)+(1)/(n_2)}}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom and the pooled variance
`S^2_p` is given by this formula:

`S^2_p =((n_1-1)S^2_1 +(n_2 -1)S^2_2)/(n_1 +n_2 -2)`

This last one is an unbiased estimator of the common variance
`\sigma^2`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 = \mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 = 0`

Alternative hypothesis:
`\mu_1 -\mu_2 \\eq 0`

Our notation on this case :

`n_1 =8` represent the sample size for group 1

`n_2 =7` represent the sample size for group 2

`\bar X_1 =19` represent the sample mean for the group 1

`\bar X_2 =22` represent the sample mean for the group 2

`s_1=3.117` represent the sample standard deviation for group 1

`s_2=5.0` represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

`S^2_p =((8-1)(3.117)^2 +(7 -1)(5.0)^2)/(8 +7 -2)=16.770`

And the deviation would be just the square root of the variance:

`S_p=4.095`

And now we can calculate the statistic:

`t=\frac{19 -22)-(0)}{4.095\sqrt{(1)/(8)+(1)/(7)}}=-1.416`

Now we can calculate the degrees of freedom given by:

`df=8+7-2=13`

And now we can calculate the p value using the altenative hypothesis:

`p_v =2*P(t_(13)<-1.416) =0.1803`

So with the p value obtained and using the significance level assumed
`\alpha=0.1` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.