Question

A random sample of size n = 48 is taken from a population with mean = -9.1 and standard deviation σ = 2. [You may find it useful to reference the z table.]

a. Calculate the expected value and the standard error for the sampling distribution of the sample mean. (Negative values should be indicated by a minus sign. Round "expected value" to 1 decimal place and "standard error" to 4 decimal places.)

Answer 1

The expected value of the sampling distribution of the sample mean is approximately -9.1, and the standard error is approximately 0.2885.

The expected value (mean) of the sampling distribution of the sample mean (often denoted as
`\mu_{\bar{x}}`) is equal to the population mean (μ). The standard error of the sample mean (
`S E_{\bar{x}}` ) is calculated using the formula:

`S E_{\bar{x}}=(\sigma)/(√(n))`

where:

`\sigma` is the population standard deviation,

n is the sample size.

Given that μ=−9.1, σ=2, and n=48, we can calculate the expected value and the standard error:

Expected Value
`\left(\mu_{\bar{x}}\right)=\mu=-9.1`

Standard Error
`\left(S E_{\bar{x}}\right)=(\sigma)/(√(n))=(2)/(√(48))`

Let's calculate the values:

Expected Value
`\left(\mu_{\bar{x}}\right)=-9.1`

Standard Error
`\left(S E_{\bar{x}}\right)=(2)/(√(48)) \approx 0.2887`

Rounded to the specified decimal places:

Expected Value
`\left(\mu_{\bar{x}}\right) \approx-9.1`

Standard Error
`\left(S E_{\bar{x}}\right) \approx 0.2887`

Answer 2

The expected value for the sampling distribution of the sample mean is -9.1 and the standard error is 0.2887

Using the parameters given for our Calculation;

• μ = -9.1
• σ = 2
• sample size, n = 48

According to a normal distribution postulate for sample sizes, n > 30. The mean of the sampling distribution is the same as the population mean. Hence, the mean of the sampling distribution is -9.1

The standard error :

• σ/√n

Now we have ;

Standard error = 2/√48 = 0.2887

Hence, the standard error is 0.2887