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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

The volume of water in a tank is 0.7 cubic meter. Write this volume in cubic centimeter

Crazy boy [7]

Answer:

11.47 in cubic centimeters

Step-by-step explanation:

6 0

3 years ago

At a fixed operating setting, the pressure in a line downstream of a reciprocating compressor has a mean value of 950 kPa with a

prisoha [69]

Answer:

4.75% probability that the line pressure will exceed 1000 kPa during any measurement

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 950, \sigma = 30$