A random sample of n = 1,000 observations from a binomial population contained 378 successes. You wish to show that p < 0.4. n = 1,000 and x = 378. You wish to show that p < 0.4. n USE SALT Calculate the appropriate test statistic. (Round your answer to two decimal places.) z = Calculate the p-value. (Round your answer to four decimal places.) p-value = Do the conclusions based on a fixed rejection region of z < -1.645 agree with those found using the p-value approach at a = 0.05? O Yes, both approaches produce the same conclusion. O No, the p-value approach rejects the null hypothesis when the fixed rejection region approach fails to reject the null hypothesis. O No, the fixed rejection region approach rejects the null hypothesis when the p-value approach fails to reject the null hypothesis. Should they? O Yes

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A random sample of \( n = 1,000 \) observations from a binomial population contained 378 successes. You wish to show that \( p < 0.4 \).

\[ n = 1,000 \text{ and } x = 378. \text{ You wish to show that } p < 0.4. \]

**Calculate the appropriate test statistic.** (Round your answer to two decimal places.)

\[ z = \underline{\hspace{2cm}} \]

**Calculate the \( p \)-value.** (Round your answer to four decimal places.)

\[ p\text{-value} = \underline{\hspace{3cm}} \]

**Do the conclusions based on a fixed rejection region of \( z < -1.645 \) agree with those found using the \( p \)-value approach at \( \alpha = 0.05 \)?**

- ○ Yes, both approaches produce the same conclusion.
- ○ No, the \( p \)-value approach rejects the null hypothesis when the fixed rejection region approach fails to reject the null hypothesis.
- ○ No, the fixed rejection region approach rejects the null hypothesis when the \( p \)-value approach fails to reject the null hypothesis.

**Should they?**

- ○ Yes
- ○ No

You may need to use the appropriate appendix table to answer this question.

---

**Explanation for Educational Purposes:**

This exercise is a hypothesis test to determine if the proportion \( p \) from a binomial distribution is less than 0.4. The sample size is 1,000, with 378 successes observed. To solve this, the test statistic \( z \) and the \( p \)-value must be calculated. Based on the \( z \)-score, the decision to reject or not reject the null hypothesis is made by comparing it to a fixed rejection region or by using the \( p \)-value method. This study aims to see if these two approaches give the same conclusion at the significance level \( \alpha = 0.05 \).
Transcribed Image Text:A random sample of \( n = 1,000 \) observations from a binomial population contained 378 successes. You wish to show that \( p < 0.4 \). \[ n = 1,000 \text{ and } x = 378. \text{ You wish to show that } p < 0.4. \] **Calculate the appropriate test statistic.** (Round your answer to two decimal places.) \[ z = \underline{\hspace{2cm}} \] **Calculate the \( p \)-value.** (Round your answer to four decimal places.) \[ p\text{-value} = \underline{\hspace{3cm}} \] **Do the conclusions based on a fixed rejection region of \( z < -1.645 \) agree with those found using the \( p \)-value approach at \( \alpha = 0.05 \)?** - ○ Yes, both approaches produce the same conclusion. - ○ No, the \( p \)-value approach rejects the null hypothesis when the fixed rejection region approach fails to reject the null hypothesis. - ○ No, the fixed rejection region approach rejects the null hypothesis when the \( p \)-value approach fails to reject the null hypothesis. **Should they?** - ○ Yes - ○ No You may need to use the appropriate appendix table to answer this question. --- **Explanation for Educational Purposes:** This exercise is a hypothesis test to determine if the proportion \( p \) from a binomial distribution is less than 0.4. The sample size is 1,000, with 378 successes observed. To solve this, the test statistic \( z \) and the \( p \)-value must be calculated. Based on the \( z \)-score, the decision to reject or not reject the null hypothesis is made by comparing it to a fixed rejection region or by using the \( p \)-value method. This study aims to see if these two approaches give the same conclusion at the significance level \( \alpha = 0.05 \).
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