Question

Consider two populations for which μ1 = 35, σ1 = 3, μ2 = 20, and σ2 = 4. Suppose that two independent random samples of sizes n1 = 47 and n2 = 54 are selected. Describe the approximate sampling distribution of x1 − x2 (center, spread, and shape). What is the shape of the distribution? The distribution would be non-normal. The distribution is approximately normal. The shape cannot be determined. What is the mean of the distribution? What is the standard deviation of the distribution?

Answer 1

The distribution is approximately normal.

The mean and standard deviation are 15 and 0.98 respectively.

Explanation:

The population of the random variables X₁ and X₂ are distributed as follows:

`X_(1)\sim (\mu_(1)=35, \sigma_(1)^(2)=3^(2))\\\\X_(2)\sim (\mu_(2)=20, \sigma_(2)^(2)=4^(2))`

Two independent random samples of sizes,

`n_(1)=47\\\\n_(2)=54`

are selected form the two populations.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample means is given by,

`\mu_(\bar x)=\mu`

And the standard deviation of the sample means is given by,

`\sigma_(\bar x)=(\sigma)/(√(n))`

Both the sample selected from the two populations are quite large, i.e.
`n_(1)=47>30\\\\n_(2)=54>30`

So, according to the central limit theorem the sampling distribution of sample means
`\bar X_(1)\ \text{and}\ \bar X_(2)` can be approximated by the Normal distribution.

Then, the distribution of
`\bar X_(1)` is:

`\bar X_(1)\sim N(35,\ 0.44)`

And the distribution of
`\bar X_(2)` is:

`\bar X_(2)\sim N(20,\ 0.54)`

If two random variables are normally distributed then their linear function is also normally distributed.

So, the distribution of
`\bar X_(1)\ - \bar X_(2)` is Normal and the shape of the distribution is bell-shaped.

The mean of the distribution of
`\bar X_(1)\ - \bar X_(2)` is:

`E(\bar X_(1)\ -\ \bar X_(2))=E(\bar X_(1))-E(\bar X_(2))\\=35-20\\=15`

The standard deviation of the distribution of
`\bar X_(1)\ - \bar X_(2)` is:

`SD(\bar X_(1)-\bar X_(2))=SD(\bar X_(1))+SD(\bar X_(2))\\=0.44+0.54\\=0.98`

*X₁ and X₂ are independent.

Thus, the mean and standard deviation are 15 and 0.98 respectively.