Question

A sociologist studying fertility in France and Switzerland wanted to test if there was a difference in the average number of babies women in each country have. The sociologist obtained a random sample of women from each country. Here are the results of their test:
Country Sample Size n Mean SD France 100 1.85 1.3 Switzerland 100 1.65 1.2
At the alfa-.05 level of significance, is there sufficient evidence to conclude that there is a difference in the average number of babies women in each country have?
a. Yes, test statistics z- 1.13 and p-vaue is.125
b. No, Test statistics z= 1.13 and p-value is .258
c. Yes, the test statistic Z- 1.13 and p-valve is 258
d. Yes, test statistics T 1.13 and p-value is .258

Answer 1

Answer:

b.

Step-by-step explanation:

From the given information:

Null and alternative hypothesis is:

$H_o : \mu _{france} = \mu_{switzerland} \\ \\ H_a : \mu _{france}\neq \mu_{switzerland}$

Given:

Number of babies   France    Switzerland

Sample size (n)         100                100

Mean                         1.85               1.65

SD                              1.3                 1.2

The test statistics can be computed as:

$Z = \dfrac{\bar x_{france} - \bar x_{switzerland}}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$

$Z = \dfrac{1.85- 1.65}{\sqrt{\dfrac{1.3^2}{100}+\dfrac{1.2^2}{100}}}$

$Z = \dfrac{0.2}{\sqrt{0.0313}}$

$Z = 1.130$

degree of freedom = $n_1 + n_2 -2$

= 100 + 100 -2

= 200 -2

=198

At 0.05, using the $t_{dist}$ table;

P-value = $t_{dist} (1.130,198,2)$

P-value = 0.258

Since P-value is greater than ∝ = 0.05, We fail to reject $H_o$

We conclude that there is No statistical difference at the 0.05 level. Hence $\mu_{france} \neq \mu_{switzerland}$