Question

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 41%. a pediatrician who works with several hospitals in the community would like to verify the hospitals claim. in a random sample of 165 babies born in the community, 78 weighed over 7 pounds. is there enough evidence to reject the hospitals claim at the 0.05 level of significance? perform a two - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis $h_0$ and the alternative hypothesis $h_1$. $h_0$: $h_1$: (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) (d) find the $p$-value. (round to three or more decimal places.) (e) can we reject the claim that the proportion of full - term babies born in the community that weigh more than 7 pounds is 41%? yes no

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0$ is the claim of the hospital. The alternative hypothesis $H_1$ for a two - tailed test is the opposite.
$H_0:p = 0.41$
$H_1:p
eq0.41$

Step2: Determine the test statistic type

We are testing a hypothesis about a proportion, so we use the z - test statistic for a proportion.

Step3: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 78$ and $n=165$. So $\hat{p}=\frac{78}{165}\approx0.47273$.

Step4: Calculate the z - test statistic

The formula for the z - test statistic for a proportion is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Substitute $p = 0.41$, $\hat{p}=0.47273$, and $n = 165$ into the formula.
$z=\frac{0.47273 - 0.41}{\sqrt{\frac{0.41\times(1 - 0.41)}{165}}}\approx\frac{0.06273}{\sqrt{\frac{0.41\times0.59}{165}}}\approx\frac{0.06273}{\sqrt{\frac{0.2419}{165}}}\approx\frac{0.06273}{\sqrt{0.00146606}}\approx\frac{0.06273}{0.03829}\approx1.638$

Step5: Calculate the p - value

For a two - tailed test with $z\approx1.638$, the p - value is $2\times(1 - P(Z<1.638))$. Using a standard normal table or calculator, $P(Z < 1.638)\approx0.9495$, so the p - value is $2\times(1 - 0.9495)=2\times0.0505 = 0.101$.

Step6: Make a decision

Since the p - value ($0.101$) is greater than the significance level $\alpha=0.05$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:p = 0.41$, $H_1:p
eq0.41$
(b) z - test statistic for a proportion
(c) $1.638$
(d) $0.101$
(e) No