To solve this problem, we can perform a z-test for a population proportion. The formula for the z-test statistic is:

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}$

Where:

• $\hat{p}$ is the sample proportion.
• $p$ is the population proportion.
• $n$ is the sample size.

Step-by-step solution:

1. Given values:

   • $p = 0.68$ (population proportion)
   • $n = 200$ (sample size)
   • $\hat{p} = \frac{140}{200} = 0.70$ (sample proportion)

2. Calculate the standard error (SE):

$SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.68 \times (1 - 0.68)}{200}} = \sqrt{\frac{0.68 \times 0.32}{200}} = \sqrt{\frac{0.2176}{200}} = \sqrt{0.001088} = 0.03299$

3. Calculate the z-test statistic:

$z = \frac{0.70 - 0.68}{0.03299} = \frac{0.02}{0.03299} \approx 0.61$

Answer:

The value of the z-test statistic is approximately 0.61, so the correct answer is D.) 0.61.

Would you like more details on this process or how hypothesis testing works?

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Related questions:

1. What is the significance of the z-test statistic in hypothesis testing?
2. How would the z-test statistic change if the sample size were larger?
3. Can you explain how to interpret the results of a z-test for proportions?
4. What is the difference between a z-test and a t-test?
5. What assumptions must be met for the z-test for a population proportion to be valid?

Tip:

In hypothesis testing, always check whether your sample size is large enough to approximate the sampling distribution as normal, especially when using the z-test for proportions.