Answer :
To answer this question, let's perform a hypothesis test to compare the true save percentages of the home and road goalies using the given information.
### Step-by-Step Solution:
#### 1. Define the Hypotheses
The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_a\)[/tex]) are:
[tex]\[ H_0: p_1 = p_2 \][/tex]
[tex]\[ H_a: p_1 > p_2 \][/tex]
where [tex]\( p_1 \)[/tex] is the true save percentage of the home goalie and [tex]\( p_2 \)[/tex] is the true save percentage of the road goalie. This is a right-tailed test.
#### 2. Collect and Summarize the Data
- Home goalie:
- Shots faced: 150
- Shots saved: 143
- Save percentage: [tex]\( \frac{143}{150} = 0.9533 \)[/tex]
- Road goalie:
- Shots faced: 144
- Shots saved: 127
- Save percentage: [tex]\( \frac{127}{144} = 0.8819 \)[/tex]
#### 3. Calculate the Pooled Proportion
The pooled proportion ([tex]\( \hat{p} \)[/tex]) is calculated by combining the total number of saves and total number of shots from both goalies:
[tex]\[ \hat{p} = \frac{\text{saves}_{\text{home}} + \text{saves}_{\text{road}}}{\text{shots}_{\text{home}} + \text{shots}_{\text{road}}} = \frac{143 + 127}{150 + 144} = \frac{270}{294} = 0.9184 \][/tex]
#### 4. Calculate the Standard Error
The standard error [tex]\( SE \)[/tex] of the difference in proportions is given by:
[tex]\[ SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \][/tex]
where [tex]\( n_1 = 150 \)[/tex] and [tex]\( n_2 = 144 \)[/tex]:
[tex]\[ SE = \sqrt{0.9184 (1 - 0.9184) \left( \frac{1}{150} + \frac{1}{144} \right)} = 0.0319 \][/tex]
#### 5. Calculate the Test Statistic
The test statistic [tex]\( z \)[/tex] is given by:
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.9533 - 0.8819}{0.0319} = 2.235 \][/tex]
#### 6. Determine the P-value
The calculated test statistic corresponds to a [tex]\( p \)[/tex]-value which can be obtained from the standard normal distribution. For a right-tailed test, the [tex]\( p \)[/tex]-value is:
[tex]\[ p \text{-value} = 1 - \Phi(z) \][/tex]
For [tex]\( z = 2.235 \)[/tex], we find:
[tex]\[ p \text{-value} = 0.013 \][/tex]
#### 7. Compare the P-value to the Significance Level
We compare the [tex]\( p \)[/tex]-value to the provided level of significance [tex]\( \alpha = 0.10 \)[/tex]:
- [tex]\( p \text{-value} = 0.013 \)[/tex]
- [tex]\( \alpha = 0.10 \)[/tex]
Since the [tex]\( p \)[/tex]-value [tex]\( (0.013) \)[/tex] is less than [tex]\( \alpha \)[/tex] [tex]\( (0.10) \)[/tex], we reject the null hypothesis.
### Conclusion
With a [tex]\( p \)[/tex]-value of 0.013, we reject the null hypothesis at the 0.10 level of significance. Therefore, we have sufficient evidence to conclude that the home goalie has a better true save percentage than the road goalie.
### Step-by-Step Solution:
#### 1. Define the Hypotheses
The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_a\)[/tex]) are:
[tex]\[ H_0: p_1 = p_2 \][/tex]
[tex]\[ H_a: p_1 > p_2 \][/tex]
where [tex]\( p_1 \)[/tex] is the true save percentage of the home goalie and [tex]\( p_2 \)[/tex] is the true save percentage of the road goalie. This is a right-tailed test.
#### 2. Collect and Summarize the Data
- Home goalie:
- Shots faced: 150
- Shots saved: 143
- Save percentage: [tex]\( \frac{143}{150} = 0.9533 \)[/tex]
- Road goalie:
- Shots faced: 144
- Shots saved: 127
- Save percentage: [tex]\( \frac{127}{144} = 0.8819 \)[/tex]
#### 3. Calculate the Pooled Proportion
The pooled proportion ([tex]\( \hat{p} \)[/tex]) is calculated by combining the total number of saves and total number of shots from both goalies:
[tex]\[ \hat{p} = \frac{\text{saves}_{\text{home}} + \text{saves}_{\text{road}}}{\text{shots}_{\text{home}} + \text{shots}_{\text{road}}} = \frac{143 + 127}{150 + 144} = \frac{270}{294} = 0.9184 \][/tex]
#### 4. Calculate the Standard Error
The standard error [tex]\( SE \)[/tex] of the difference in proportions is given by:
[tex]\[ SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \][/tex]
where [tex]\( n_1 = 150 \)[/tex] and [tex]\( n_2 = 144 \)[/tex]:
[tex]\[ SE = \sqrt{0.9184 (1 - 0.9184) \left( \frac{1}{150} + \frac{1}{144} \right)} = 0.0319 \][/tex]
#### 5. Calculate the Test Statistic
The test statistic [tex]\( z \)[/tex] is given by:
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.9533 - 0.8819}{0.0319} = 2.235 \][/tex]
#### 6. Determine the P-value
The calculated test statistic corresponds to a [tex]\( p \)[/tex]-value which can be obtained from the standard normal distribution. For a right-tailed test, the [tex]\( p \)[/tex]-value is:
[tex]\[ p \text{-value} = 1 - \Phi(z) \][/tex]
For [tex]\( z = 2.235 \)[/tex], we find:
[tex]\[ p \text{-value} = 0.013 \][/tex]
#### 7. Compare the P-value to the Significance Level
We compare the [tex]\( p \)[/tex]-value to the provided level of significance [tex]\( \alpha = 0.10 \)[/tex]:
- [tex]\( p \text{-value} = 0.013 \)[/tex]
- [tex]\( \alpha = 0.10 \)[/tex]
Since the [tex]\( p \)[/tex]-value [tex]\( (0.013) \)[/tex] is less than [tex]\( \alpha \)[/tex] [tex]\( (0.10) \)[/tex], we reject the null hypothesis.
### Conclusion
With a [tex]\( p \)[/tex]-value of 0.013, we reject the null hypothesis at the 0.10 level of significance. Therefore, we have sufficient evidence to conclude that the home goalie has a better true save percentage than the road goalie.
