The question in the image involves selecting the appropriate test statistic and calculating it for two scenarios. Here's the solution:

Part (a):

Given:

• Sample size $n = 11$,
• Population is normally distributed,
• Population standard deviation ($\sigma$) is known: $2.3$.

Since the population standard deviation is known, use the $z$-statistic.

The formula for $z$ is: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ Here:

• $\bar{x} = 29.6$ (sample mean),
• $\mu = 30.3$ (population mean),
• $\sigma = 2.3$,
• $n = 11$.

Now calculate: $z = \frac{29.6 - 30.3}{\frac{2.3}{\sqrt{11}}} = \frac{-0.7}{\frac{2.3}{3.3166}} = \frac{-0.7}{0.6934} \approx -1.01$

Answer for (a): $z = -1.01$.

Part (b):

Given:

• Sample size $n = 20$,
• Population distribution is unknown.

Since the population distribution is unknown and the sample size is relatively small ($<30$), use the $t$-statistic.

The formula for $t$ is: $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ Here:

• $\bar{x} = 29.6$,
• $\mu = 30.3$,
• $s = 2.3$ (sample standard deviation),
• $n = 20$.

Now calculate: $t = \frac{29.6 - 30.3}{\frac{2.3}{\sqrt{20}}} = \frac{-0.7}{\frac{2.3}{4.4721}} = \frac{-0.7}{0.5145} \approx -1.36$

Answer for (b): $t = -1.36$.

Summary of Answers:

• (a) $z = -1.01$,
• (b) $t = -1.36$.

Do you need further clarifications or detailed steps?
Here are 5 questions to deepen your understanding:

1. Why do we use $z$-statistics when the population standard deviation is known?
2. What are the critical differences between $z$-statistics and $t$-statistics?
3. How does the sample size influence the decision to use $z$ or $t$?
4. What assumptions are required for using the $t$-statistic in small samples?
5. How would the test statistics change if the sample sizes were larger?

Tip: Always verify if the population standard deviation or sample standard deviation is provided—it directly affects the choice of test statistic!