**Educational Resource on Statistical Analysis****Statistical Problem Solving: Medical Hay Fever Rates**To analyze the incidence of hay fever in specific regions, we examine data from two different sample groups. This involves calculating statistics such as sample means, variances, and testing hypotheses about population parameters.**Data Overview**1. **Regions Studied:** - **New England (Sample $n_1$):** Data on the hay fever rate for people under 25. - **Western Kansas (Sample $n_2$):** Data on the hay fever rate for people over 50.2. **Hay Fever Rates:** - **Under 25 (sample $n_1$):** The rates per 1000 population are 132, 112, 127, 93, 123, 122, 120. - **Over 50 (sample $n_2$):** The rates per 1000 population are 128, 101, 97, 120, 125, 98, 95.**Statistical Goals**- **Hypothesis Testing:** - **Problem 1:** Compare the average hay fever rates between these two groups. Use a significance level ($\alpha = 0.05$) to identify if there is a statistically significant difference.**Graphical Representation**No graphs are included in the data provided. The analysis can include visualizations like bar graphs or box plots to illustrate the distribution and mean differences of hay fever rates between the two age groups.---This resource provides essential practice in hypothesis testing and interpreting real-world health data, ideal for students and educators in statistics and health sciences.

Name: **Educational Resource on Statistical Analysis****Statistical Problem
Uploaded: 2024-03-20T07:07:00.000Z
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Description: **Educational Resource on Statistical Analysis****Statistical Problem Solving: Medical Hay Fever Rates**To analyze the incidence of hay fever in specific regions, we examine data from two different sample groups. This involves calculating statistics

i. Mean and standard deviation for : Use Ti 83 calculator to find the mean and standard deviation as follows: Select STAT > Edit > Enter the values of . Click 2nd button; take the arrow to the MATH menu, and then ‘3’ numbered key. Click 2nd button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket. Click Enter. Click 2nd button; take the arrow to the MATH menu, and then ‘7’ numbered key. Click 2nd button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket. Click Enter. Output using Ti 83 calculator is given below: From the Ti 83 calculator output, the mean value is 109.50, and standard deviation value is 15.41. Mean and standard deviation for : Use Ti 83 calculator to find the mean and standard deviation as follows: Select STAT > Edit > Enter the values of . Click 2nd button; take the arrow to the MATH menu, and then ‘3’ numbered key. Click 2nd button; then ‘2’ numbered key to get L2, and close the ‘)’ bracket. Click Enter. Click 2nd button; take the arrow to the MATH menu, and then ‘7’ numbered key. Click 2nd button; then ‘2’ numbered key to get L2, and close the ‘)’ bracket. Click Enter. Output using Ti 83 calculator is given below: From the Ti 83 calculator output, the mean value is 99.36, and standard deviation value is 11.57. Hence, the sample mean and the sample standard deviation by using the calculator is verified.ii. Let μ1 denotes the mean Hay fever of the age group under 25, μ2 denotes the mean Hay fever of the age group over 50. The claim of the test is that, mean Hay fever of the age group under 25 is higher than the mean Hay fever of the age group over 50. Null hypothesis: H0 : μ1=μ2 Alternative hypothesis: Ha : μ1>μ2Degrees of freedom: The degrees of freedom for the t distribution is, df=smallern1-1, n2-1 In the formula n1 is the sample size from population 1 and n2 is the sample size from population 2. Substitute as 16, as 14 in the degrees of freedom formula df=smaller16-1, 14-1=smaller15, 13=13 The degrees of freedom are 13.The distribution of hay fever rate in each age group is approximately normal distribution and the population standard deviation is unknown. Hence, the sampling distribution to be used is Student’s t distribution. The two sample t statistic is, t=x¯1-x¯2-μ1-μ2s12n1+s22n2=109.5-99.36-015.41216+11.57214=10.144.9400=2.053 Hence, the sample test statistic is 2.053.Use Ti 83 calculator to obtain the P-value as follows Select STAT, take the arrow to the TESTS menu, and then ‘4’ numbered key. Select Stats under Inpt. Enter x¯1 as 5, Sx1 as 15.41, and as 16. Enter x¯2 as 36, Sx2 as 11.57, and as 14. Select the option >μ2. Select No under Pooled. Click Draw. Output using Ti 83 calculator is given below: From the Ti 83 calculator output, the P-value is 0.0249.Rejection rule: If the P-value is less than or equal to , then reject the null hypothesis and the test is statistically significant. That is, p-value≤α. Conclusion: The P-value is 0.0249 and the level of significance is 0.05. The P-value is less than the level of significance. By the rejection rule, the null hypothesis is rejected and the data is statistically significant at level 0.05. The null hypothesis is rejected. This shows that, there is sufficient evidence that the mean Hay fever of the age group under 25 is higher than the mean Hay fever of the age group over 50 at level of significance 0.05.

Math

Statistics

Medical: Hay Fever A random sample of n, = 16 communities in western Kansas gave the following information for people under 25 years of age. x,: Rate of hay fever per 1000 population for people under 25 120 128 92 123 112 93 86 06 125 95 125 117 97 122 127 88 Arandom sample of n2 information for people over 50 years old. = 14 regions in western Kansas gave the following x,: Rate of hay fever per 1000 population for people over 50 25 101 97 112 88 115 114 85 68 96 (Reference: National Center for Health Statistics.) i. Use a calculator to verify that x 109.50, s, 15.41, 99.36, anc S2 = 11.57. ii. Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has lower rate of hay fever? Use a 0.05.

Medical: Hay Fever A random sample of n, = 16 communities in western Kansas gave the following information for people under 25 years of age. x,: Rate of hay fever per 1000 population for people under 25 120 128 92 123 112 93 86 06 125 95 125 117 97 122 127 88 Arandom sample of n2 information for people over 50 years old. = 14 regions in western Kansas gave the following x,: Rate of hay fever per 1000 population for people over 50 25 101 97 112 88 115 114 85 68 96 (Reference: National Center for Health Statistics.) i. Use a calculator to verify that x 109.50, s, 15.41, 99.36, anc S2 = 11.57. ii. Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has lower rate of hay fever? Use a 0.05.

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Concept explainers

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.

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Question

**Educational Resource on Statistical Analysis**

**Statistical Problem Solving: Medical Hay Fever Rates**

To analyze the incidence of hay fever in specific regions, we examine data from two different sample groups. This involves calculating statistics such as sample means, variances, and testing hypotheses about population parameters.

**Data Overview**

1. **Regions Studied:**
- **New England (Sample $n_1$):** Data on the hay fever rate for people under 25.
- **Western Kansas (Sample $n_2$):** Data on the hay fever rate for people over 50.

2. **Hay Fever Rates:**
- **Under 25 (sample $n_1$):** The rates per 1000 population are 132, 112, 127, 93, 123, 122, 120.
- **Over 50 (sample $n_2$):** The rates per 1000 population are 128, 101, 97, 120, 125, 98, 95.

**Statistical Goals**

- **Hypothesis Testing:**
- **Problem 1:** Compare the average hay fever rates between these two groups. Use a significance level ($\alpha = 0.05$) to identify if there is a statistically significant difference.

**Graphical Representation**

No graphs are included in the data provided. The analysis can include visualizations like bar graphs or box plots to illustrate the distribution and mean differences of hay fever rates between the two age groups.

---

This resource provides essential practice in hypothesis testing and interpreting real-world health data, ideal for students and educators in statistics and health sciences.

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