The product of r and s
2 answers:
You might be interested in
First, compute the mean: 
Next compute the absolute deviations for each datapoint:
|75-95.555|=20.555
|89-95.555|=6.555
|145-95.555|=49.445
|85-95.555|=10.555
|80-95.555|=15.555
|92-95.555|=3.555
|104-95.555|=8.445
|90-95.555|=5.555
|100-95.555|=4.445
Next we'll take the mean of these deviations:
The mean absolute deviation of your data is 13.8517
Next compute the absolute deviations for each datapoint:
|75-95.555|=20.555
|89-95.555|=6.555
|145-95.555|=49.445
|85-95.555|=10.555
|80-95.555|=15.555
|92-95.555|=3.555
|104-95.555|=8.445
|90-95.555|=5.555
|100-95.555|=4.445
Next we'll take the mean of these deviations:
The mean absolute deviation of your data is 13.8517
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:
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
The standard error of the difference between means is
Then, the statistic can be calculated as:
The P-value for this z-statistic in a two tailed test is:
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.
Answer:
Hope this will help you. Here you have to use the formula P = 2(l+b)
