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

Answer:

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

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:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

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

Step-by-step 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:

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

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:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

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: $\mu_1 \neq \mu_2$

Or equivalently:

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

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

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:

$S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770$

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

$S_p=4.095$

And now we can calculate the statistic:

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

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:

$p_v =2*P(t_{13}$

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.