An airline company is interested in improving customer satisfaction rate from the 50% currently claimed. The company sponsored a survey of 244 customers and found that 130 customers were satisfied. Determine whether sufficient evidence exists that the customer satisfaction rate is higher than the claim by the company. What is the test statistic ?What is the -value?Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level of ?Round  to two decimal places.Round the -value to three decimal places.

To determine whether sufficient evidence exists that the customer satisfaction rate is higher than the claimed 50%, we can conduct a hypothesis test for the population proportion. Here's how we go about it step by step:

### Step 1: Set up the hypotheses

- **Null hypothesis \(H_0\):** The customer satisfaction rate is equal to the company's claimed rate. This means \( p = 0.50 \).
- **Alternative hypothesis \(H_1\):** The customer satisfaction rate is higher than the claimed rate. This means \( p > 0.50 \).

This is a **one-tailed test** since we are only interested in whether the satisfaction rate is higher than the claimed rate.

### Step 2: Compute the sample proportion

The sample proportion \( \hat{p} \) is calculated as follows:

\[
\hat{p} = \frac{\text{number of satisfied customers}}{\text{total number of customers in the sample}} = \frac{130}{244} = 0.5328
\]

### Step 3: Calculate the test statistic

We use the formula for the test statistic for a proportion:

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

Where:
- \( \hat{p} = 0.5328 \) is the sample proportion,
- \( p_0 = 0.50 \) is the claimed proportion,
- \( n = 244 \) is the sample size.

Plugging in the values:

\[
z = \frac{0.5328 - 0.50}{\sqrt{\frac{0.50(1 - 0.50)}{244}}}
= \frac{0.0328}{\sqrt{\frac{0.25}{244}}}
= \frac{0.0328}{0.0319}
= 1.28
\]

The test statistic is approximately \( z = 1.28 \).

### Step 4: Calculate the p-value

Since this is a one-tailed test, the p-value corresponds to the area to the right of \( z = 1.28 \) in the standard normal distribution.

Using standard normal distribution tables or a calculator:

\[
\text{p-value} = P(Z > 1.28) = 0.1003
\]

The p-value is approximately \( 0.100 \).

### Step 5: Make a decision

Let's assume the significance level \( \alpha = 0.05 \).

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

Since the p-value of 0.100 is greater than the significance level of 0.05, we **fail to reject the null hypothesis**. This means that there is **not sufficient evidence** to conclude that the customer satisfaction rate is higher than the claimed 50%.

### Conclusion

- **Test statistic (z-value):** \( 1.28 \)
- **p-value:** \( 0.100 \)
- **Conclusion:** At a 0.05 significance level, there is not enough evidence to suggest that the customer satisfaction rate is higher than 50%.

Would you like more details or have any questions?

Here are 5 related questions that build on this:

1. What would happen if the sample size was larger?
2. How does changing the significance level affect the conclusion?
3. What is the difference between a one-tailed and two-tailed hypothesis test?
4. How is the p-value interpreted in the context of hypothesis testing?
5. Can you explain what Type I and Type II errors are in hypothesis testing?

**Tip:** In hypothesis testing, always check if your test is one-tailed or two-tailed, as it affects how you calculate the p-value.

An airline company is interested in improving customer satisfaction rate from the 50% currently claimed. The company sponsored a survey of 244 customers and found that 130 customers were satisfied. Determine whether sufficient evidence exists that the customer satisfaction rate is higher than the claim by the company. What is the test statistic? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level of 0.05? Round the test statistic to two decimal places. Round the p-value to three decimal places.

Determine whether the customer satisfaction rate of an airline exceeds the claimed 50%. Conduct a hypothesis test for a population proportion using sample data from 244 customers, where 130 were satisfied. Find the test statistic, p-value, and determine if there's sufficient evidence to reject the null hypothesis at a 0.05 significance level.

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