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Anestetic [448]
2 years ago
6

Plzzzz helpppppppp...........

Mathematics
1 answer:
Flura [38]2 years ago
8 0

Answer:

Why isn't there a picture-

Step-by-step explanation:

HEHEH BOIS

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Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio
pychu [463]

Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

4. Do not reject H0

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.

5 0
3 years ago
Simplify: <br> (8)2(2)3<br> (4)3(-2)4
Lostsunrise [7]
1). (16)(6)

2). (12)(-8)
8 0
3 years ago
Help me with please i will give a brainliest for a good answer
viva [34]
4(2) - 5 = 3
4(5) - 5 = 15
4(8) - 5 = 27
4(9) - 5 = 31

so the answers are 3, 15, 27, and 31
6 0
3 years ago
Read 2 more answers
Suppose the number of free throws in a basketball game by one player are normally distributed with a standard deviation 0.97 fre
babunello [35]

Answer: 1.960

Step-by-step explanation:

The value of z we use to calculate a confidence interval with a (1-\alpha) confidence level is a two-tailed test value i.e. represented by :-

              z_{\alpha/2}

Given : The level of confidence: 1-\alpha=0.95

Then, significance level : \alpha: 1-0.95=0.05

With the help of standard normal distribution table for z , we have

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.960

Hence, the value of z should be used to calculate a confidence interval with a 95% confidence level =1.960

3 0
3 years ago
How do you learn pi? What is it used for? Please be specific when explaining.
Ugo [173]

Answer:

just memorize the numbers in pi, I know the first like 4 numbers. 3.1415

Step-by-step explanation:

it is used for calculating the area of circles.

4 0
2 years ago
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