Questions: A random sample of single males and single females (ages 21 to 76) was obtained. The sample includes those currently separated, divorced, or widowed. The subjects were given a trait and asked to respond whether the trait was a deal-breaker in terms of maintaining the relationship. The results of the survey are provided in the table for the trait "stubborn". Is there a significant difference between males and females who consider the trait "stubborn" a relationship deal-breaker? Use the alpha=0.05 level of significance. Determine the size of each sample. nM=546 nF=554 Determine the sample proportion of each gender that considers stubbornness to be a relationship deal-breaker. p̂M=square p̂F=square

A random sample of single males and single females (ages 21 to 76) was obtained. The sample includes those currently separated, divorced, or widowed. The subjects were given a trait and asked to respond whether the trait was a deal-breaker in terms of maintaining the relationship. The results of the survey are provided in the table for the trait "stubborn". Is there a significant difference between males and females who consider the trait "stubborn" a relationship deal-breaker? Use the alpha=0.05 level of significance. Determine the size of each sample. nM=546 nF=554 Determine the sample proportion of each gender that considers stubbornness to be a relationship deal-breaker. p̂M=square p̂F=square
Transcript text: A random sample of single males and single females (ages 21 to 76 ) was obtained. The sample includes those currently separated, divorced, or widowed. The subjects were given a trait and asked to respond whether the trait was a deal-breaker in terms of maintaining the relationship. The results of the survey are provided in the table for the trait "stubborn". Is there a significant difference between males and females who consider the trait "stubborn" a relationship deal-breaker? Use the $\alpha=0.05$ level of significance. Determine the size of each sample. \[ \begin{array}{l} n_{M}=546 \\ n_{F}=554 \end{array} \] Determine the sample proportion of each gender that considers stubbornness to be a relationship deal-breaker. \[ \begin{array}{l} \hat{\mathrm{p}}_{\mathrm{M}}=\square \\ \hat{\mathrm{p}}_{\mathrm{F}}=\square \end{array} \]
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Solution

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Solution Steps

Step 1: Sample Sizes and Proportions

The sample sizes for the two groups are: \[ n_{M} = 546 \quad \text{(males)} \] \[ n_{F} = 554 \quad \text{(females)} \]

The sample proportions of each gender that consider stubbornness a deal-breaker are calculated as follows: \[ \hat{p}_{M} = \frac{228}{546} \approx 0.418 \] \[ \hat{p}_{F} = \frac{312}{554} \approx 0.563 \]

Step 2: Expected Frequencies

The expected frequencies for each cell in the contingency table are calculated using the formula: \[ E = \frac{R_i \times C_j}{N} \] where \(R_i\) is the total for row \(i\), \(C_j\) is the total for column \(j\), and \(N\) is the total sample size.

Calculating the expected frequencies:

  • For cell (1, 1): \[ E = \frac{546 \times 540}{1100} \approx 268.0364 \]
  • For cell (1, 2): \[ E = \frac{546 \times 560}{1100} \approx 277.9636 \]
  • For cell (2, 1): \[ E = \frac{554 \times 540}{1100} \approx 271.9636 \]
  • For cell (2, 2): \[ E = \frac{554 \times 560}{1100} \approx 282.0364 \]

The expected frequencies are: \[ \begin{bmatrix} 268.0364 & 277.9636 \\ 271.9636 & 282.0364 \end{bmatrix} \]

Step 3: Chi-Square Test Statistic

The Chi-Square test statistic (\(\chi^2\)) is calculated using the formula: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \] where \(O\) is the observed frequency and \(E\) is the expected frequency.

Calculating the contributions to the Chi-Square statistic:

  • For cell (1, 1): \[ O = 228, \quad E \approx 268.0364 \quad \Rightarrow \quad \frac{(228 - 268.0364)^2}{268.0364} \approx 5.9802 \]
  • For cell (1, 2): \[ O = 318, \quad E \approx 277.9636 \quad \Rightarrow \quad \frac{(318 - 277.9636)^2}{277.9636} \approx 5.7666 \]
  • For cell (2, 1): \[ O = 312, \quad E \approx 271.9636 \quad \Rightarrow \quad \frac{(312 - 271.9636)^2}{271.9636} \approx 5.8938 \]
  • For cell (2, 2): \[ O = 242, \quad E \approx 282.0364 \quad \Rightarrow \quad \frac{(242 - 282.0364)^2}{282.0364} \approx 5.6833 \]

Summing these contributions gives: \[ \chi^2 \approx 22.7451 \]

Step 4: Critical Value and P-Value

The critical value for a Chi-Square distribution with 1 degree of freedom at \(\alpha = 0.05\) is: \[ \chi^2_{\alpha, df} = \chi^2_{(0.05, 1)} \approx 3.8415 \]

The p-value associated with the calculated Chi-Square statistic is: \[ P = P(\chi^2 > 22.7451) = 0.0 \]

Final Answer

Since the calculated Chi-Square statistic \(22.7451\) exceeds the critical value \(3.8415\) and the p-value is \(0.0\), we reject the null hypothesis. There is a significant difference between males and females who consider the trait "stubborn" a relationship deal-breaker.

\[ \boxed{\text{There is a significant difference between males and females regarding stubbornness as a deal-breaker.}} \]

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