Question

The Central Limit Theorem allows us to estimate the parameters as well as describe the distribution for a sampling distribution. Which of the following descriptions is false? O If N is large, we can compute the standard error of the mean using a specified formula O None of these are false O If N is large, the mean of the sampling distribution is the same as the mean of the population from which the samples were selected. If N is large, the sampling distribution of the mean will be normally distributed, no matter what the shape of the population is.

Answer 1

1.the central limit theorem (CLT) states that the distribution of sample means approximates a normal distribution (also known as a “bell curve”), as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population distribution shape.

Hence option(D) is correct.

2.

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1 (Xˉ1​) =	18.56
Sample Standard Deviation 1 (s1​) =	4.35
Sample Size 1 (N1​) =	43
Sample Mean 2 (Xˉ2​) =	17.95
Sample Standard Deviation 2 (s_2)(s2​) =	4.87
Sample Size 2 (N_2)(N2​) =	38

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df=n1​+n2​−2=43+38−2=79.

The critical value for \alpha = 0.05α=0.05 and df = 79df=79 degrees of freedom is $t_c = t_{1-\alpha/2; n-1} = 1.99$ The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

$\begin{array}{ccl} s_p = \displaystyle \sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} }= \displaystyle \sqrt{ \frac{(43-1)\times 4.35^2 + (38-1) \times 4.87^2}{43 + 38 - 2} }=4.601 \end{array}$

Since we assume that the population variances are equal, the standard error is computed as follows:

$\begin{array}{ccl} se =\displaystyle s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } =\displaystyle 4.601 \times \sqrt{ \frac{1}{43} + \frac{1}{38}}= 1.024 \end{array}$

Now, we finally compute the confidence interval:

$\begin{array}{ccl} CI=\displaystyle \left( \bar X_1 - \bar X_2 - t_c \times se, \bar X_1 - \bar X_2 + t_c \times se \right)= \displaystyle \left( 18.56 - 17.95 - 1.99 \times 1.024, 18.56 - 17.95 + 1.99 \times 1.024 \right) = (-1.429, 2.649) \end{array}$CI​=(−1.429,2.649)​

Therefore, based on the data provided, the 95\%95% confidence interval for the difference between the population means μ1​−μ2​ is −1.429<μ1​−μ2​<2.649, which indicates that we are 95% confident that the true difference between population means is contained by the interval (-1.429, 2.649)

please rate my answer and comment for doubts.