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Sagot :
Answer:
b.
Step-by-step explanation:
From the given information:
Null and alternative hypothesis is:
[tex]H_o : \mu _{france} = \mu_{switzerland} \\ \\ H_a : \mu _{france}\neq \mu_{switzerland}[/tex]
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:
[tex]Z = \dfrac{\bar x_{france} - \bar x_{switzerland}}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}[/tex]
[tex]Z = \dfrac{1.85- 1.65}{\sqrt{\dfrac{1.3^2}{100}+\dfrac{1.2^2}{100}}}[/tex]
[tex]Z = \dfrac{0.2}{\sqrt{0.0313}}[/tex]
[tex]Z = 1.130[/tex]
degree of freedom = [tex]n_1 + n_2 -2[/tex]
= 100 + 100 -2
= 200 -2
=198
At 0.05, using the [tex]t_{dist}[/tex] table;
P-value = [tex]t_{dist} (1.130,198,2)[/tex]
P-value = 0.258
Since P-value is greater than ∝ = 0.05, We fail to reject [tex]H_o[/tex]
We conclude that there is No statistical difference at the 0.05 level. Hence [tex]\mu_{france} \neq \mu_{switzerland}[/tex]
The significance level shows that there is no sufficient evidence to conclude that there is a difference in the average number of babies as B. Test statistics z= 1.13 and p-value is 0.258
What is a significance level?
It should be noted that a significance level simply means the probability that an event could have taken place by chance.
In this case, the test statistic will be:
= (1.85 - 1.65)/(✓1.3²/100 + 1.2²/100)
= 1.13
This is a two tailed test. Therefore, the p value will be:
= 2(1 - 0.87076)
= 0.258
Therefore, the p value is more than 0.05 and there's no sufficient evidence to reject the null hypothesis.
Learn more about significance level on:
https://brainly.com/question/4599596
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