Question

Perform these steps. a. Find the Spearman rank correlation coefficient. b. State the hypotheses. c. Find the critical value. Use \(\alpha=0.05\). d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Hospitals and Nursing Homes Find the Spearman rank correlation coefficient for the following data, which represent the number of hospitals and nursing homes in each of seven randomly selected states. At the 0.05 level of significance, is there enough evidence to conclude that there is a correlation between the two? $$ \begin{array}{l|ccccccc} \text { Hospitals } & 107 & 61 & 202 & 133 & 145 & 117 & 108 \\ \hline \text { Nursing homes } & 230 & 134 & 704 & 376 & 431 & 538 & 373 \end{array} $$

Short answer

At 0.05 significance level, there is no evidence of correlation.

Step-by-step solution

Rank the Data

First, rank the hospitals and nursing homes data separately. Assign ranks from 1 to 7 based on the ordered numbers, with the smallest number getting rank 1. In case of ties, assign the average rank for the tied values. **Hospitals Ranks**: 2, 1, 7, 4, 6, 3, 5 **Nursing Homes Ranks**: 2, 1, 7, 4, 5, 6, 3.

Calculate Differences and Squares of Ranks

Calculate the difference between the ranks of hospitals and nursing homes for each state and then square those differences.Let \( d_i = r_{i,\text{Hospitals}} - r_{i,\text{Nursing Homes}} \).- \( d_1 = 2 - 2 = 0 \), \( d_1^2 = 0^2 = 0 \)- \( d_2 = 1 - 1 = 0 \), \( d_2^2 = 0^2 = 0 \)- \( d_3 = 7 - 7 = 0 \), \( d_3^2 = 0^2 = 0 \)- \( d_4 = 4 - 4 = 0 \), \( d_4^2 = 0^2 = 0 \)- \( d_5 = 6 - 5 = 1 \), \( d_5^2 = 1^2 = 1 \)- \( d_6 = 3 - 6 = -3 \), \( d_6^2 = (-3)^2 = 9 \)- \( d_7 = 5 - 3 = 2 \), \( d_7^2 = 2^2 = 4 \)Sum of squared differences: \( \sum d_i^2 = 0 + 0 + 0 + 0 + 1 + 9 + 4 = 14 \).

Calculate Spearman Rank Correlation Coefficient

Use the formula for the Spearman rank correlation coefficient:\[ r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)} \]where \( n \) is the number of data pairs (here, 7).Substitute the values:\[ r_s = 1 - \frac{6 \times 14}{7 \times (7^2 - 1)} = 1 - \frac{84}{7 \times 48} = 1 - \frac{84}{336} = 1 - 0.25 = 0.75 \].

State the Hypotheses

For hypothesis testing of Spearman's rank correlation coefficient, the hypotheses are:- Null hypothesis \( H_0 \): There is no correlation between hospitals and nursing homes; \( \rho_s = 0 \).- Alternative hypothesis \( H_a \): There is a correlation between hospitals and nursing homes; \( \rho_s eq 0 \).

Find the Critical Value

For a significance level \( \alpha = 0.05 \), a two-tailed test (since the hypothesis is \( \rho_s eq 0 \)), find the critical value for \( n = 7 \) from Spearman's rank correlation critical value table. For \( n = 7 \), the critical value at \( \alpha = 0.05 \) is approximately \( \pm0.786 \).

Make the Decision

Compare the calculated Spearman rank correlation coefficient \( r_s = 0.75 \) with the critical value \( \pm0.786 \).Since \( 0.75 \) is less than \( 0.786 \) and does not meet or exceed critical value levels, fail to reject the null hypothesis.

Summarize the Results

At the 0.05 level of significance, there is not enough evidence to conclude that there is a correlation between the number of hospitals and nursing homes in the selected states.

Key concepts

Hypothesis Testing

Hypothesis testing is a powerful statistical tool used to make decisions about data. In this context, it helps to determine if there is a significant correlation between two variables, like hospitals and nursing homes in this exercise. The process begins by stating the hypotheses.

• Null Hypothesis \( H_0 \): This represents the default position that there is no relationship or correlation, in this case, between the number of hospitals and nursing homes.
• Alternative Hypothesis \( H_a \): This suggests that there is a relationship or correlation between the two variables.

Spearman's rank correlation test helps to ascertain these hypotheses using ranks of the data, rather than raw values, making it perfect for non-parametric data. Once the hypotheses are stated, the test proceeds to determine if the correlation found is significant enough to reject the null hypothesis, leaning towards the alternative hypothesis.

Critical Values

Critical values are thresholds that define the boundaries of acceptance or rejection areas for a hypothesis test. These values are determined based on the significance level \( \alpha \) and the distribution of the test statistic. For Spearman's rank correlation, these values can be found in statistical tables.

In this exercise, the significance level is set at \( \alpha = 0.05 \). For a two-tailed test involving Spearman's rank correlation with 7 data points, the critical value is approximately \( \pm0.786 \). This means that if the calculated Spearman rank correlation coefficient falls within \(-0.786\) and \(0.786\), the null hypothesis cannot be rejected.

• If the test statistic exceeds these critical limits, the null hypothesis is rejected.
• If the test statistic is within these limits, there isn’t enough evidence to reject the null hypothesis.

Identifying and using critical values correctly is a key step in ensuring accurate results in hypothesis testing.

Correlation Coefficient

The correlation coefficient is a statistical measure that indicates the strength and direction of a relationship between two variables. In Spearman's rank correlation, the correlation coefficient \( r_s \) is used to gauge how well the relationship between two variables can be described by a monotonic function.

The correlation coefficient is calculated as follows:\[ r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}\]where \( d_i \) are the differences between the ranks of the two variables for each data pair, and \( n \) is the total number of data pairs.

• If \( r_s = 1 \), perfect positive correlation exists.
• If \( r_s = -1 \), perfect negative correlation exists.
• If \( r_s = 0 \), no correlation exists.

In the exercise, the computed \( r_s = 0.75 \) suggests a moderate positive correlation. However, it does not lie beyond the critical value of \( \pm0.786 \), thus there is not enough evidence to assert a statistically significant correlation.

Non-parametric Statistics

Non-parametric statistics are methods that do not assume a specific distribution for the dataset. These methods are especially useful when data do not meet the assumptions necessary for parametric tests, such as normality.

Spearman's rank correlation is an example of a non-parametric test. It works by ranking data points rather than using their raw values, making it less sensitive to outliers and ideal for ordinal data or data that do not meet the assumptions of correlation measures like Pearson's coefficient.

• Advantages: No assumption of normality; useful for ordinal data.
• Disadvantages: May be less powerful than parametric tests when assumptions are met.

Using non-parametric statistics extends the ability to analyze data that might otherwise be unsuitable for traditional methods, as seen in the exercise where ranks allowed the exploration of a potential relationship between hospitals and nursing homes.