We are conducting a hypothesis test of the claim that the population mean cholesterol level for middle-aged adults is less than 186.6 mg/dL. The sample data provided is:

• Sample mean: $\bar{x} = 190.6$ mg/dL
• Population claim (null hypothesis mean): $\mu = 186.6$ mg/dL
• Sample standard deviation: $s = 18$ mg/dL

Let's address each scenario:

(a) The sample has size 10, and it is from a population with a distribution about which we know very little.

Since we have a small sample size ($n = 10$) and we know little about the population distribution, we cannot assume normality or use the population standard deviation. Thus, we should use the t-statistic. The formula for the t-statistic is:

$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{x}$ is the sample mean,
• $\mu$ is the population mean (under the null hypothesis),
• $s$ is the sample standard deviation,
• $n$ is the sample size.

Let's calculate the t-statistic:

$t = \frac{190.6 - 186.6}{\frac{18}{\sqrt{10}}}$

First, calculate the standard error:

$\text{Standard error} = \frac{18}{\sqrt{10}} = \frac{18}{3.162} \approx 5.69$

Now, calculate the t-statistic:

$t = \frac{190.6 - 186.6}{5.69} = \frac{4}{5.69} \approx 0.70$

So, the t-statistic for this scenario is:

$t \approx 0.70$

(b) The sample has size 100, and it is from a non-normally distributed population with a known standard deviation.

Since we have a large sample size ($n = 100$) and a known standard deviation (which we assume to be the population standard deviation $\sigma$), we can use the z-statistic. The central limit theorem tells us that with a large sample size, the sampling distribution of the sample mean will be approximately normal, regardless of the population distribution. The formula for the z-statistic is:

$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

However, you did not provide the population standard deviation $\sigma$. Let's assume that $\sigma = 18$ mg/dL (the same as the sample standard deviation for simplicity), but if this value is different, you would need to replace it with the correct population standard deviation.

Using this assumption, let's calculate the z-statistic:

$z = \frac{190.6 - 186.6}{\frac{18}{\sqrt{100}}}$

First, calculate the standard error:

$\text{Standard error} = \frac{18}{\sqrt{100}} = \frac{18}{10} = 1.8$

Now, calculate the z-statistic:

$z = \frac{190.6 - 186.6}{1.8} = \frac{4}{1.8} \approx 2.22$

So, the z-statistic for this scenario is:

$z \approx 2.22$

Summary:

• (a) t-statistic = 0.70
• (b) z-statistic = 2.22

Do you need further clarification on any of these steps?

Related Questions:

1. How does the choice of test statistic change if the sample size is increased?
2. What happens if we do not know the population standard deviation in large samples?
3. What are the assumptions behind using the t-distribution in hypothesis testing?
4. Why is the central limit theorem important in hypothesis testing?
5. How would you determine the critical values for the t- or z-statistic for a given significance level?

Tip:

For small samples, especially when the population distribution is unknown, always use the t-statistic unless the population standard deviation is known and the data are normally distributed.