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A random sample of \(n_{1}=16\) communities in western Kansas gave the following information for people under 25 years of age. \(x_{1}:\) Rate of hay fever per 1000 population for people under 25 \(\begin{array}{rrrrrrrr}98 & 90 & 120 & 128 & 92 & 123 & 112 & 93 \\ 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88\end{array}\) A random sample of \(n_{2}=14\) regions in western Kansas gave the following information for people over 50 years old. \(x_{2}\) : Rate of hay fever per 1000 population for people over 50 \(\begin{array}{llllllr}95 & 110 & 101 & 97 & 112 & 88 & 110 \\ 79 & 115 & 100 & 89 & 114 & 85 & 96\end{array}\) (Reference: National Center for Health Statistics.) i. Use a calculator to verify that \(\bar{x}_{1} \approx 109.50, s_{1} \approx 15.41, \bar{x}_{2} \approx 99.36\), and \(s_{2} \approx 11.57\) 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 a lower rate of hay fever? Use \(\alpha=0.05\).

Step by step solution

01

State Hypotheses

We need to test if people over 50 years have a lower mean rate of hay fever than those under 25. Set up the hypothesis as follows: - Null Hypothesis: \(H_0: \mu_1 \leq \mu_2\) (Mean rate for under 25 is less than or equal to over 50)- Alternative Hypothesis: \(H_a: \mu_1 > \mu_2\) (Mean rate for under 25 is greater than over 50)

02

Determine Test Statistic

We will use a t-test for the difference of means, assuming unequal variances:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values: \[ t = \frac{109.50 - 99.36}{\sqrt{\frac{15.41^2}{16} + \frac{11.57^2}{14}}} \]

03

Calculate Test Statistic

Calculate the standard error:\[ \sqrt{\frac{15.41^2}{16} + \frac{11.57^2}{14}} \approx 5.735 \]Then, calculate the t value:\[ t \approx \frac{109.50 - 99.36}{5.735} \approx 1.77 \]

04

Determine Degrees of Freedom

Use the formula for degrees of freedom (df) with unequal variances:\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]Plug in values to calculate df:\[ df \approx 25 \] (using nearest integer for critical values lookup)

05

Find Critical Value and Decision

For \(\alpha = 0.05\), the critical t-value for a one-tailed test with 25 degrees of freedom is approximately 1.708.Since our computed t-statistic (1.77) is greater than 1.708, we reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing

Let's start with hypothesis testing, a foundational concept in statistics. It helps determine if there's enough evidence in our sample data to infer that a certain condition holds for the entire population.
The process involves comparing two hypotheses:

• Null Hypothesis (\(H_0\)): This is a statement suggesting no effect or no difference exist. In our exercise, it suggests that the rate of hay fever for people under 25 is less than or equal to those over 50.
• Alternative Hypothesis (\(H_a\)): This contradicts the null hypothesis. It proposes that people under 25 have a higher rate of hay fever than those over 50.

After establishing these hypotheses, the next step is to determine if the data provides enough evidence to reject the null hypothesis. We set a significance level (commonly \(\alpha = 0.05\)) to minimize incorrect decisions.

Degrees of Freedom

Degrees of freedom can be a tricky concept, but it’s crucial for accurate hypothesis testing. It refers to the number of independent values that can vary in our data analysis.
In a t-test for two independent samples with unequal variances, the degrees of freedom are computed using a slightly complex formula considering sample sizes and variances. \[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]Using the given sample and their variances, we approximated 25 degrees of freedom. This value is critical as it guides us to the appropriate "critical value" when consulting statistical t-distribution tables.

Critical Value

The critical value helps in deciding whether to reject the null hypothesis. For a given level of significance (\(\alpha\)), it acts as a threshold.
In the context of our example, we perform a one-tailed test (since we hypothesize one rate is greater than the other) with a \(\alpha = 0.05\). Given the 25 degrees of freedom, our critical value is found in t-distribution tables as approximately 1.708.
The decision rule is simple: if our calculated t-statistic exceeds this critical value, reject the null hypothesis. With a calculated t-statistic of 1.77, which exceeds 1.708, the data suggests evidence strong enough to reject the null hypothesis.

Standard Error

Understanding standard error is essential when comparing means between two groups. It quantifies the amount of variability or dispersion from the mean in our sampling.
In our exercise, we calculated it using the formula:\[\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Plugging in our given sample variances and sizes, it provided an approximate value of 5.735.
This value represents the standard deviation of the sampling distribution of the sample mean differences. The smaller this value, the more accurate our estimates of the population means are. It directly affects the computation of the t-statistic and influences the strength of our conclusions in hypothesis testing.