Class problem $$ \# 10 $$ proetice: Michael decided to enroll in a reading course that allegedly increases reading speed. Prior to enrolling in the class, Michael read 198 words per minute (wpm). The following data represent the words per minute read for 10 different passages read after the course: $$ 206 \quad 217 \quad 197 \quad 199 \quad 210 \quad 210 \quad 197 \quad 212 \quad 227 \quad 209 $$ $$ \begin{array}{l}\text { At the } \alpha=0.1 \text { level of significance, determine the P-value and use it to decide if the } \ \text { class was effective. }\end{array} $$

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To determine whether Michael's reading course was effective in increasing his reading speed, we'll perform a hypothesis test using the provided data.

### **Step 1: Define the Hypotheses**
- **Null Hypothesis ($$H_0$$)**: The course has no effect on Michael's reading speed.
  $$
  H_0: \mu = 198 \text{ wpm}
  $$
- **Alternative Hypothesis ($$H_1$$)**: The course has increased Michael's reading speed.
  $$
  H_1: \mu > 198 \text{ wpm}
  $$

### **Step 2: Calculate the Sample Mean ($$\bar{x}$$)**
The given reading speeds after the course are:
$$
206, \ 217, \ 197, \ 199, \ 210, \ 210, \ 197, \ 212, \ 227, \ 209
$$

$$
\bar{x} = \frac{206 + 217 + 197 + 199 + 210 + 210 + 197 + 212 + 227 + 209}{10} = \frac{2084}{10} = 208.4 \text{ wpm}
$$

### **Step 3: Calculate the Sample Standard Deviation ($$s$$)**
First, find each deviation from the mean, square it, and sum them:

$$
\begin{align*}
(206 - 208.4)^2 &= 5.76 \
(217 - 208.4)^2 &= 73.96 \
(197 - 208.4)^2 &= 129.96 \
(199 - 208.4)^2 &= 88.36 \
(210 - 208.4)^2 &= 2.56 \
(210 - 208.4)^2 &= 2.56 \
(197 - 208.4)^2 &= 129.96 \
(212 - 208.4)^2 &= 12.96 \
(227 - 208.4)^2 &= 345.96 \
(209 - 208.4)^2 &= 0.36 \
\hline
\text{Sum of Squared Deviations} &= 792.4
\end{align*}
$$

$$
s^2 = \frac{792.4}{9} \approx 88.04 \quad \Rightarrow \quad s \approx \sqrt{88.04} \approx 9.38 \text{ wpm}
$$

### **Step 4: Compute the Test Statistic ($$t$$)**
$$
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{208.4 - 198}{9.38 / \sqrt{10}} = \frac{10.4}{2.968} \approx 3.498
$$

### **Step 5: Determine the P-value**
Using the t-distribution with $$df = n - 1 = 9$$:

A $$t$$-value of **3.498** exceeds the critical value for $$df = 9$$ at $$ \alpha = 0.005 $$. Therefore, the P-value is **less than 0.005**.

For more precision, using a t-distribution calculator:
$$
\text{P-value} \approx 0.002 \quad (\text{one-tailed})
$$

### **Step 6: Make a Decision**
Compare the P-value to the significance level ($$\alpha = 0.1$$):
$$
0.002 < 0.1
$$
Since the P-value is less than $$\alpha$$, we **reject the null hypothesis**.

### **Conclusion**
There is sufficient evidence at the 0.1 significance level to conclude that the reading course was effective in increasing Michael's reading speed.

**Summary:**
- **P-value ≈ 0.002**
- **Decision:** Reject $$H_0$$; the class was effective.
The P-value is approximately 0.002, which is less than the significance level of 0.1. Therefore, we reject the null hypothesis and conclude that the reading course was effective in increasing Michael's reading speed.

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