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beks73 [17]
3 years ago
5

The table below shows the values of y for different values of x:

Mathematics
1 answer:
Jobisdone [24]3 years ago
3 0
Y = 3x

Putting (0,0)
the equation is satisfied.

Putting (1,3)
3 = 3(1) , satisfied

Putting (2,6)
6 = 3(2), satisfied

Putting (3,9)
9 = 3(3), satisfied
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Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

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Step-by-step explanation:

We have to perfomr an hypothesis test to see if there is strong evidence that there is a significant difference between the two population means.

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq0

Being μ1 the mean average gain for younger mothers and μ2 the mean average gain for mature mothers.

(NOTE: we are comparing means, not proportions, as it is the random variable is the weight gain).

As we are claiming "strong evidence", the level of significance will be 0.01.

For younger mothers, the sample size is n1=840, the sample mean is 30.7 and  the sample standard deviation is s1=14.91.

For mature mothers, the sample size is n2=132, the sample mean is 29.15 and the sample standard deviation is s2=13.46.

The difference between means is

M_d=\mu_1-\mu_2=30.7-29.15=1.55

The standard error of the difference between means is

s_M=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{14.91^2}{840}+\dfrac{13.46^2}{132}}=\sqrt{ 0.2647+1.3725}=\sqrt{1.6372}\\\\\\s_M=1.2795

Then, the statistic can be calculated as:

z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{1.55-0}{1.2795}=1.2114

The P-value for this z-statistic in a two tailed test is:

P-value=2P(z>1.2114)=0.2257

As the P-value is greater than the significance level, the null hypothesis failed to be rejected.

There is no enough evidence to claim that the real average weight gain differs from mature and youger mothers.

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