Based on their records, a hospital claims that the proportion, *p*, of full-term babies born in the community that weigh more than 7 pounds is 44%. A pediatrician who works with several hospitals in the community would like to verify the hospital's claim. In a random sample of 210 babies born in the community, 99 weighed over 7 pounds. Is there enough evidence to reject the hospital's claim at the 0.05 level of significance?

Perform a two-tailed test. 

Carry your intermediate computations to three or more decimal places.

To determine whether there is enough evidence to reject the hospital's claim at the 0.05 level of significance, we can perform a two-tailed hypothesis test for the population proportion. Here's the step-by-step process:

### Step 1: State the Hypotheses
- **Null Hypothesis (\(H_0\))**: The proportion of full-term babies that weigh more than 7 pounds is 0.44.
  \[
  H_0: p = 0.44
  \]
- **Alternative Hypothesis (\(H_1\))**: The proportion of full-term babies that weigh more than 7 pounds is not 0.44.
  \[
  H_1: p \neq 0.44
  \]

### Step 2: Calculate the Sample Proportion
The sample proportion (\(\hat{p}\)) is calculated as:
\[
\hat{p} = \frac{\text{Number of babies weighing over 7 pounds}}{\text{Total number of babies in the sample}} = \frac{99}{210} \approx 0.4714
\]

### Step 3: Calculate the Test Statistic
The test statistic for a proportion is calculated using the formula:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \(\hat{p} = 0.4714\) (sample proportion)
- \(p_0 = 0.44\) (claimed proportion)
- \(n = 210\) (sample size)

Substituting the values:
\[
z = \frac{0.4714 - 0.44}{\sqrt{\frac{0.44(1 - 0.44)}{210}}} = \frac{0.0314}{\sqrt{\frac{0.44 \times 0.56}{210}}} = \frac{0.0314}{\sqrt{\frac{0.2464}{210}}} = \frac{0.0314}{\sqrt{0.001173}} = \frac{0.0314}{0.0342} \approx 0.918
\]

### Step 4: Determine the Critical Value and P-value
For a two-tailed test at the 0.05 level of significance, the critical z-values are \( \pm 1.96 \).

Now, find the p-value corresponding to the calculated \( z \)-value. The p-value is the probability of observing a test statistic at least as extreme as the one computed, assuming that the null hypothesis is true.

Using the standard normal distribution table or a calculator, the p-value for \( z = 0.918 \) is approximately \( 0.359 \) for one tail, so the two-tailed p-value is \( 2 \times 0.359 = 0.718 \).

### Step 5: Make a Decision
- If the p-value is less than 0.05, reject the null hypothesis.
- If the p-value is greater than 0.05, fail to reject the null hypothesis.

In this case, \( 0.718 \) is much greater than 0.05, so we **fail to reject the null hypothesis**.

### Conclusion
There is not enough evidence at the 0.05 level of significance to reject the hospital's claim that the proportion of full-term babies born in the community that weigh more than 7 pounds is 44%.

Would you like more details or have any further questions? Here are some related questions:

1. How would the conclusion change if the sample size were larger?
2. What are the implications of failing to reject the null hypothesis in this context?
3. How do you interpret the p-value in the context of this hypothesis test?
4. What if the level of significance was 0.01 instead of 0.05?
5. How does the power of the test affect the likelihood of detecting a false null hypothesis?

**Tip:** Always ensure your sample size is large enough to meet the conditions for using a normal approximation in hypothesis testing for proportions.

Learn how to perform hypothesis testing to verify a hospital's claim about the proportion of full-term babies weighing over 7 pounds using a Z-test. In a sample of 210 babies, 99 weighed over 7 pounds, leading to a calculated test statistic and p-value. Determine significance at the 0.05 level.

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