Mathematics, 13.07.2021 15:10 chrismcb875

Consider the following results for two independent random samples taken from two populations. Sample 1 Sample 2
n1=50 n2=35
x¯1=13.6 x¯2=11.6
σ1=2.2 σ2=3.0

Required:
a. What is the point estimate of the difference between the two population means?
b. Provide a 90% confidence interval for the difference between the two population means.
c. Provide a 95% confidence interval for the difference between the two population means.

Answers: 1

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Answer from: BARRION1981

a. 2

b. The 90% confidence interval for the difference between the two population means is (1.02, 2.98).

c. The 95% confidence interval for the difference between the two population means is (0.83, 3.17).

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes 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 between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample 1:

$\mu_1 = 13.6, s_1 = \frac{2.2}{\sqrt{50}} = 0.3111$

Sample 2:

$\mu_2 = 11.6, s_2 = \frac{3}{\sqrt{35}} = 0.5071$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 13.6 - 11.6 = 2$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.3111^2+0.5071^2} = 0.595$

a. What is the point estimate of the difference between the two population means?

Sample difference, so $\mu = 2$

b. Provide a 90% confidence interval for the difference between the two population means.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of 1 - 0.05 = 0.95 , so Z = 1.645.

The margin of error is:

M = zs = 1.645(0.595) = 0.98

The lower end of the interval is the sample mean subtracted by M. So it is 2 - 0.98 = 1.02

The upper end of the interval is the sample mean added to M. So it is 2 + 0.98 = 2.98

The 90% confidence interval for the difference between the two population means is (1.02, 2.98).

c. Provide a 95% confidence interval for the difference between the two population means.

Following the same logic as b., we have that Z = 1.96 . So

M = zs = 1.96(0.595) = 1.17

The lower end of the interval is the sample mean subtracted by M. So it is 2 - 1.17 = 0.83

The upper end of the interval is the sample mean added to M. So it is 2 + 1.17 = 3.17

The 95% confidence interval for the difference between the two population means is (0.83, 3.17).

Answer from: Quest

there will be 35 boys in each team. : )

Answer from: Quest

7.666270253x10 raised to the 18th power

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