SOLUTION: An article published in the Washington Post claims that 45 percent of all Americans have brown eyes. A random sample of n=76 college students students found 28 who had brown eyes.

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Question 1169896: An article published in the Washington Post claims that 45 percent of all Americans have brown eyes. A random sample of n=76 college students students found 28 who had brown eyes.
H0:p=.45
Ha:p≠.45
(a) The test statistic is z =
(b) P-value =
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down this hypothesis testing problem step by step.
**Understanding the Problem**
We're conducting a two-tailed z-test for proportions to determine if the proportion of college students with brown eyes is significantly different from the claimed 45% of all Americans.
**Given Information**
* Claimed population proportion (p0) = 0.45
* Sample size (n) = 76
* Number of college students with brown eyes (x) = 28
* Sample proportion (p̂) = x/n = 28/76 ≈ 0.3684
* Null hypothesis (H0): p = 0.45
* Alternative hypothesis (Ha): p ≠ 0.45
**(a) Calculating the Test Statistic (z)**
The formula for the z-test statistic for proportions is:
$$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$$
Where:
* p̂ is the sample proportion (0.3684)
* p0 is the hypothesized population proportion (0.45)
* n is the sample size (76)
Let's plug in the values:
$$z = \frac{0.3684 - 0.45}{\sqrt{\frac{0.45(1 - 0.45)}{76}}}$$
$$z = \frac{-0.0816}{\sqrt{\frac{0.45(0.55)}{76}}}$$
$$z = \frac{-0.0816}{\sqrt{\frac{0.2475}{76}}}$$
$$z = \frac{-0.0816}{\sqrt{0.0032565789}}$$
$$z = \frac{-0.0816}{0.057066442}$$
$$z \approx -1.4299$$
Therefore, the test statistic is approximately z = -1.43.
**(b) Calculating the p-value**
Since this is a two-tailed test (Ha: p ≠ 0.45), we need to find the area in both tails of the standard normal distribution.
1. **Find the area to the left of z = -1.43:**
* Using a standard normal distribution table or a calculator, the area to the left of z = -1.43 is approximately 0.0764.
2. **Double the area for the two-tailed p-value:**
* Since it's a two-tailed test, we multiply the area by 2:
* p-value = 2 * 0.0764 = 0.1528.
Therefore, the p-value is approximately 0.1528.
**Answers**
(a) The test statistic is z ≈ -1.43.
(b) The p-value ≈ 0.1528.