Question

) Study cited that the calcium levels in men and women ages 70 years or older are summarized that sample size of 55 men with mean 785 and standard deviation 489 and sample size of 40 women with mean 670 and standard deviation of 419, what is the probability of obtaining a difference between sample means of 110 mg or more?

Answer 1

Answer:

0.5188 = 51.88% probability of obtaining a difference between sample means of 110 mg or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution, the central limit theorem and difference between normal variables.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal sample means:

When we subtract two distribution of the sample means, which are normal, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of standard errors squared.

Sample size of 55 men with mean 785 and standard deviation 489

This means that:

$\mu_A = 785$

$s_A = \frac{489}{\sqrt{55}} = 65.9367$

Sample size of 40 women with mean 670 and standard deviation of 419

This means that:

$\mu_B = 670$

$s_B = \frac{419}{\sqrt{40}} = 66.2497$

Difference between sample means:

The mean and the standard deviation are, respectively:

$\mu = \mu_A - \mu_B = 785 - 670 = 115$

$\sigma = \sqrt{s_A^2+s_B^2} = \sqrt{65.9367^2+66.2497^2} = 93.47$

What is the probability of obtaining a difference between sample means of 110 mg or more?

This is 1 subtracted by the pvalue of Z when X = 110. So

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

$Z = \frac{110 - 115}{93.47}$

$Z = -0.053$

$Z = -0.053$ has a pvalue of 0.4812.

1 - 0.4812 = 0.5188

0.5188 = 51.88% probability of obtaining a difference between sample means of 110 mg or more