To analyze whether the hospital's claim that 42% of full-term babies weigh over 7 pounds is accurate, we can perform a hypothesis test for a population proportion.

Step-by-Step Solution:

1. Sample Size (n): 175
2. Sample Proportion ($\hat{p}$) of babies over 7 pounds: $\frac{91}{175} \approx 0.520$
3. Population Proportion ($p_0$): 0.42
4. Level of Significance ($\alpha$): 0.05

We'll go through each question step by step:

(a) State the null hypothesis $H_0$ and the alternative hypothesis $H_1$

Since we are conducting a two-tailed test, we want to see if the proportion significantly differs from 42%.

• $H_0: p = 0.42$ (The proportion of babies weighing over 7 pounds is 42%)
• $H_1: p \neq 0.42$ (The proportion of babies weighing over 7 pounds is not 42%)

(b) Determine the type of test statistic to use

Since we are testing for a population proportion, we use a z-test for proportions.

(c) Find the value of the test statistic

The formula for the z-test statistic for a population proportion is:

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

Plugging in the values:

$\hat{p} = 0.520, \quad p_0 = 0.42, \quad n = 175$

$z = \frac{0.520 - 0.42}{\sqrt{\frac{0.42 \times 0.58}{175}}}$

Calculating step-by-step:

1. $p_0 \times (1 - p_0) = 0.42 \times 0.58 = 0.2436$
2. $\frac{0.2436}{175} \approx 0.001392$
3. $\sqrt{0.001392} \approx 0.0373$
4. $z = \frac{0.520 - 0.42}{0.0373} \approx 2.679$

So, the test statistic is approximately $z \approx 2.679$.

(d) Find the two critical values

For a two-tailed test with $\alpha = 0.05$, the critical z-values are based on the tails of the standard normal distribution.

Using a z-table or calculator, the critical values for a two-tailed test at $\alpha = 0.05$ are approximately:

$\pm 1.96$

(e) Can we reject the claim that the proportion of full-term babies born in the hospital that weigh more than 7 pounds is 42%?

To determine if we reject the null hypothesis, compare the test statistic to the critical values:

• If $|z| > 1.96$, we reject $H_0$.
• Here, $|z| = 2.679$, which is greater than 1.96.

Since $2.679 > 1.96$, we reject the null hypothesis. Therefore, there is enough evidence to conclude that the proportion of babies weighing over 7 pounds is significantly different from 42%.

Conclusion:

• Answer: Yes, we reject the claim that the proportion is 42%.

Would you like further details on any of these steps?

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Additional Questions for Practice:

1. How would the test change if the level of significance were 0.01 instead of 0.05?
2. If the sample size were smaller (say, 50), how might this affect the test's reliability?
3. How can we interpret the p-value in the context of this hypothesis test?
4. What would happen if the observed proportion was exactly 0.42?
5. Could a one-tailed test be appropriate in any scenario for this problem?

Tip:

In hypothesis testing, larger sample sizes generally yield more reliable results, reducing the margin of error for the estimated proportion.