Two Populations 9) To compare the performance of students in private and public schools, samples of reading test scores were collected from two groups of students. In a sample of 106 private-school students the sample mean was 282 with a standard deviation of 46 and in a sample of 151 public-school students, the sample mean was 265 with a standard deviation of 68 . a) At the $$ 1 \% $$ level of significance does it appear the scores are the same? b) What is the p-value?

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To compare the reading test scores of students from private and public schools, we'll perform a two-sample hypothesis test. Here's a step-by-step solution to both parts of the question:

### **Given Data:**
- **Private Schools:**
  - Sample size ($$n_1$$) = 106
  - Sample mean ($$\bar{x}_1$$) = 282
  - Sample standard deviation ($$s_1$$) = 46

- **Public Schools:**
  - Sample size ($$n_2$$) = 151
  - Sample mean ($$\bar{x}_2$$) = 265
  - Sample standard deviation ($$s_2$$) = 68

### **Part (a): Hypothesis Test at the 1% Significance Level**

**1. **_**State the Hypotheses:**_**
- **Null Hypothesis ($$H_0$$)**: There is no difference in mean scores between private and public schools. ($$\mu_1 = \mu_2$$)
- **Alternative Hypothesis ($$H_1$$)**: There is a difference in mean scores between private and public schools. ($$\mu_1 \neq \mu_2$$)

**2. **_**Calculate the Test Statistic:**_**
We'll use a **z-test** since both sample sizes are large.

$$
z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
$$

Plugging in the values:

$$
z = \frac{(282 - 265)}{\sqrt{\frac{46^2}{106} + \frac{68^2}{151}}} = \frac{17}{\sqrt{\frac{2116}{106} + \frac{4624}{151}}} = \frac{17}{\sqrt{19.92 + 30.63}} = \frac{17}{\sqrt{50.55}} \approx \frac{17}{7.11} \approx 2.392
$$

**3. **_**Determine the Critical Value:**_**
For a two-tailed test at the 1% significance level ($$\alpha = 0.01$$), the critical z-values are $$\pm 2.576$$.

**4. **_**Make a Decision:**_**
- If $$|z| > 2.576$$, reject $$H_0$$.
- Here, $$|2.392| < 2.576$$, so we **fail to reject** the null hypothesis.

**Conclusion for Part (a):**
At the **1% significance level**, there is **not enough evidence** to conclude that there is a difference in mean reading test scores between private and public school students.

### **Part (b): Calculating the p-value**

**1. **_**Find the p-value:**_**
The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis.

$$
p\text{-value} = 2 \times P(Z > |z|)
$$

From Part (a), $$z \approx 2.392$$.

Using standard normal distribution tables or a calculator:

$$
P(Z > 2.392) \approx 0.0083
$$

Thus,

$$
p\text{-value} \approx 2 \times 0.0083 = 0.0166 \text{ or } 1.66\%
$$

**Conclusion for Part (b):**
The **p-value is approximately 1.66%**.

**Implications:**
- Since the p-value (1.66%) is greater than the significance level ($$\alpha = 1%$$), it corroborates our decision in Part (a) to **fail to reject** the null hypothesis.
At the 1% significance level, there is not enough evidence to conclude that the mean reading test scores differ between private and public school students. The p-value is approximately 1.66%.

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