Questions: For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim. An amusement park claims that the mean daily attendence at the park is 21,000 people. Write the claim as a mathematical statement. A. µ ≠ 21,000 B. µ=21,000 C. µ<21,000 D u < 21000 E. u ≥ 21.000 F. µ>21,000

For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim.

An amusement park claims that the mean daily attendence at the park is 21,000 people.

Write the claim as a mathematical statement.
A. µ ≠ 21,000 B. µ=21,000 C. µ<21,000
D u < 21000 E. u ≥ 21.000 F. µ>21,000

Transcript text: For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim. An amusement park claims that the mean daily attendence at the park is 21,000 people. Write the claim as a mathematical statement. A. $\mu \neq 21,000$ B. $\mu=21,000$ C. $\mu<21,000$ D u < 21000 E. $u \geq 21.000$ F. $\mu>21,000$

Solution

Solution Steps

Step 1: Claim and Hypotheses

The amusement park claims that the mean daily attendance at the park is 21,000 people. We can express this claim mathematically as:

\[ \mu = 21000 \]

We set up the null and alternative hypotheses as follows:

Null Hypothesis ($H_0$): $\mu = 21000$
Alternative Hypothesis ($H_a$): $\mu \neq 21000$

Step 2: Calculate Standard Error

The standard error ($SE$) is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{300}{\sqrt{30}} \approx 54.7723 \]

Step 3: Calculate Test Statistic

The test statistic ($Z_{test}$) is calculated using the formula:

\[ Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{21050 - 21000}{54.7723} \approx 0.9129 \]

Step 4: Calculate P-value

For a two-tailed test, the p-value is calculated as:

\[ P = 2 \times (1 - T(|z|)) \approx 0.3613 \]

Step 5: Conclusion

To determine whether to reject the null hypothesis, we compare the p-value to the significance level ($\alpha = 0.05$). Since $0.3613 > 0.05$, we fail to reject the null hypothesis.

Final Answer

The null hypothesis is not rejected, indicating that there is not enough evidence to dispute the amusement park's claim that the mean daily attendance is 21,000 people.

Thus, the answer is:

\[ \boxed{B} \]

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