A poll of

1228
 
Americans showed that

46.9

​% 

of the respondents prefer to watch the news rather than read or listen to it. Use those results with a

0.10
 
significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. Use the​ P-value method. Use the normal distribution as an approximation to the binomial distribution.

Let p denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it. Identify the null and alternative hypotheses.

We are tasked with testing a claim about the proportion of Americans who prefer to watch the news rather than read or listen to it. Let's follow the steps of hypothesis testing and use the **P-value method**.

### Step 1: **Define the Null and Alternative Hypotheses**
Let \( p \) represent the population proportion of Americans who prefer to watch the news.

- **Null Hypothesis ( \( H_0 \) )**: The null hypothesis is the claim that the population proportion is at least 50%, i.e., \( p = 0.5 \).
  
  \[
  H_0: p \geq 0.5
  \]
  
- **Alternative Hypothesis ( \( H_a \) )**: The alternative hypothesis is the claim that fewer than half of Americans prefer to watch the news, i.e., \( p < 0.5 \).
  
  \[
  H_a: p < 0.5
  \]

This is a **left-tailed** test since we are testing if the proportion is less than 0.5.

### Step 2: **Calculate the Test Statistic**

We will use the normal approximation to the binomial distribution since the sample size is large. The formula for the test statistic \( z \) is:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \( \hat{p} \) is the sample proportion (46.9% or 0.469),
- \( p_0 \) is the hypothesized population proportion (0.5),
- \( n \) is the sample size (1228).

Let's compute the test statistic.

The test statistic \( z \) is approximately \(-2.17\).

### Step 3: **Find the P-value**

Since this is a left-tailed test, we need to find the probability that a standard normal variable is less than \(-2.17\). This corresponds to the P-value, which we can determine using the standard normal distribution.

Let's compute the P-value.The P-value is approximately 0.0149.

### Step 4: **Decision Rule**

We compare the P-value to the significance level \( \alpha = 0.10 \).

- If the P-value is less than \( \alpha \), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than or equal to \( \alpha \), we fail to reject \( H_0 \).

Since the P-value (0.0149) is less than the significance level of 0.10, we **reject the null hypothesis**.

### Conclusion:
There is sufficient evidence to support the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it.

Would you like more details or have any questions?

### 5 Related Questions:
1. What is the significance of using the P-value method in hypothesis testing?
2. How does the normal approximation to the binomial distribution work?
3. What would the outcome be if the significance level was set to 0.05 instead?
4. Can you explain the meaning of a left-tailed test in more depth?
5. How do you interpret a P-value in hypothesis testing?

**Tip:** Always ensure the sample size is large enough for the normal approximation to the binomial distribution to be valid (typically \( n \times p_0 \geq 5 \) and \( n \times (1 - p_0) \geq 5 \)).

A poll of 1228 Americans showed that 46.9% of the respondents prefer to watch the news rather than read or listen to it. Use those results with a 0.10 significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. Use the​ P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it. Identify the null and alternative hypotheses.

Test the claim that fewer than half of Americans prefer to watch the news with a sample of 1228 respondents. Learn how to use hypothesis testing, the P-value method, and the normal approximation to the binomial distribution. Step-by-step solution with a significance level of 0.10.

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