In a location two random samples were taken concerning the rate of hay fever per 1000 population. Sample 1 (n, = 14) contained people under 25, sample 2 (n, = 16) contained people over 50. %3D %=109.5 s, = 15.41 X2 = 99.36 s2 = 11.57 %3D Assume that the hay fever.rate in each grcup has an approximately normal distribution. Do the data indicate that the over 50 group has a significantly lower hay fever rate? (Use a = 5%) %3D

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#6 it must be in the format of the second picture.

**Title: Statistical Analysis of Hay Fever Rates**

In a certain location, researchers conducted a study to understand the rate of hay fever per 1000 population. Two random samples were taken:

- **Sample 1**: Consisted of 14 people under the age of 25.
- **Sample 2**: Consisted of 16 people over the age of 50.

**Statistics Gathered:**
- Mean of Sample 1 (\( \bar{X}_1 \)): 109.5 with a standard deviation (\( s_1 \)) of 15.41
- Mean of Sample 2 (\( \bar{X}_2 \)): 99.36 with a standard deviation (\( s_2 \)) of 11.57

**Research Question:**
Assuming that the hay fever rate in each group is approximately normally distributed, the aim is to determine whether the data indicates that the over 50 age group has a significantly lower hay fever rate. A significance level (\( \alpha \)) of 5% is used.

**Components for Analysis:**

- **Random Variable**: The rate of hay fever in each age group.
- **Special Considerations/Characteristics**: The normality condition of the data distribution.
- **Given**: Data and sample details as previously listed.

**Question to Answer:**
Does the above data support the hypothesis that people over 50 have a significantly lower rate of hay fever compared to those under 25?

**Statistical Testing Components:**
- **Testing Type**: (To be specified based on hypothesis test, e.g., t-test)
- **Justification**: Explanation needed for the type of test chosen.
- **Test Statistic**: Calculation based on data.
- **Test Statistic Value**: Numerical result to compare against critical value.

**Visual Aid Description (If applicable):**
The worksheet includes sections for specifying details of the hypothesis test, though no graphs or diagrams are present in the image.

This type of statistical analysis is critical for understanding the epidemiology of diseases like hay fever across different demographics.
Transcribed Image Text:**Title: Statistical Analysis of Hay Fever Rates** In a certain location, researchers conducted a study to understand the rate of hay fever per 1000 population. Two random samples were taken: - **Sample 1**: Consisted of 14 people under the age of 25. - **Sample 2**: Consisted of 16 people over the age of 50. **Statistics Gathered:** - Mean of Sample 1 (\( \bar{X}_1 \)): 109.5 with a standard deviation (\( s_1 \)) of 15.41 - Mean of Sample 2 (\( \bar{X}_2 \)): 99.36 with a standard deviation (\( s_2 \)) of 11.57 **Research Question:** Assuming that the hay fever rate in each group is approximately normally distributed, the aim is to determine whether the data indicates that the over 50 age group has a significantly lower hay fever rate. A significance level (\( \alpha \)) of 5% is used. **Components for Analysis:** - **Random Variable**: The rate of hay fever in each age group. - **Special Considerations/Characteristics**: The normality condition of the data distribution. - **Given**: Data and sample details as previously listed. **Question to Answer:** Does the above data support the hypothesis that people over 50 have a significantly lower rate of hay fever compared to those under 25? **Statistical Testing Components:** - **Testing Type**: (To be specified based on hypothesis test, e.g., t-test) - **Justification**: Explanation needed for the type of test chosen. - **Test Statistic**: Calculation based on data. - **Test Statistic Value**: Numerical result to compare against critical value. **Visual Aid Description (If applicable):** The worksheet includes sections for specifying details of the hypothesis test, though no graphs or diagrams are present in the image. This type of statistical analysis is critical for understanding the epidemiology of diseases like hay fever across different demographics.
**Transcription of Statistical Analysis Worksheet**

---

**Test #2**

- **Random Variable:** \(\bar{x}\)  
- **Special considerations/characteristics:**  

**Given:**  
\[ P(\bar{x} \leq 6820) = 0.0030 \]

---

**Sample:**  
- \( n = 50 \)  
- \(\bar{x} = 6820\)

**Sampling Distribution (\(\bar{x}\)):**

- **B. What Type:** Normal
- **Why Justified:** CLT n ≥ 30
- **Test Statistic:** \( z \)  
- **Test Statistic Value:**

  \[ \mu_{\bar{x}} = \mu = 7500 \]  
  \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1750}{\sqrt{50}} = 247.4874 \]

  \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{6820 - 7500}{247.4874} = -2.75 \]

**Population:**  
- \( \mu = 7500 \)  
- \( \sigma = 1750 \)

---

**C. Drawing:**

\[ P(\bar{x} \leq 6820) \]   
\[ P(z \leq -2.75) = 0.0030 \]

**Graph:**  

There is a normal distribution curve drawn, with a mean at the center. A vertical line is drawn at \( z = -2.75 \), marking the left tail of the curve. The shaded region to the left of this line represents the probability, \( 0.0030 \).

---

**Additional Notes:**

- The use of the Central Limit Theorem (CLT) is noted to justify treating the sampling distribution as approximately normal due to the sample size (\( n = 50 \)).
- Calculation steps for standard error and z-value are provided.

---
Transcribed Image Text:**Transcription of Statistical Analysis Worksheet** --- **Test #2** - **Random Variable:** \(\bar{x}\) - **Special considerations/characteristics:** **Given:** \[ P(\bar{x} \leq 6820) = 0.0030 \] --- **Sample:** - \( n = 50 \) - \(\bar{x} = 6820\) **Sampling Distribution (\(\bar{x}\)):** - **B. What Type:** Normal - **Why Justified:** CLT n ≥ 30 - **Test Statistic:** \( z \) - **Test Statistic Value:** \[ \mu_{\bar{x}} = \mu = 7500 \] \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1750}{\sqrt{50}} = 247.4874 \] \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{6820 - 7500}{247.4874} = -2.75 \] **Population:** - \( \mu = 7500 \) - \( \sigma = 1750 \) --- **C. Drawing:** \[ P(\bar{x} \leq 6820) \] \[ P(z \leq -2.75) = 0.0030 \] **Graph:** There is a normal distribution curve drawn, with a mean at the center. A vertical line is drawn at \( z = -2.75 \), marking the left tail of the curve. The shaded region to the left of this line represents the probability, \( 0.0030 \). --- **Additional Notes:** - The use of the Central Limit Theorem (CLT) is noted to justify treating the sampling distribution as approximately normal due to the sample size (\( n = 50 \)). - Calculation steps for standard error and z-value are provided. ---
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