Answer :
To address this hypothesis testing problem step-by-step, here's how we proceed:
### 1. Stating the Hypotheses:
We are given a significance level of [tex]\( \alpha = 0.05 \)[/tex]. The hypotheses for this problem are:
- Null Hypothesis [tex]\( H_0 \)[/tex]: The mean amount of time spent watching television by women is greater than or equal to the mean amount of time spent watching television by men.
[tex]\[ H_0: \mu_1 \geq \mu_2 \][/tex]
- Alternative Hypothesis [tex]\( H_a \)[/tex]: The mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.
[tex]\[ H_a: \mu_1 < \mu_2 \][/tex]
### 2. Calculating the Test Statistic:
We need to calculate the test statistic for the difference in means, assuming unequal variances. The formula for the test statistic [tex]\( t \)[/tex] is:
[tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} } \][/tex]
Given the summary statistics:
[tex]\[ \bar{x}_1 = 12.5 \, \text{hr}, \quad \bar{x}_2 = 13.8 \, \text{hr} \][/tex]
[tex]\[ s_1 = 3.9 \, \text{hr}, \quad s_2 = 5.2 \, \text{hr} \][/tex]
[tex]\[ n_1 = 14, \quad n_2 = 17 \][/tex]
First, we calculate the standard error (SE):
[tex]\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3.9^2}{14} + \frac{5.2^2}{17}} \approx 1.639 \][/tex]
Now the test statistic [tex]\( t \)[/tex] is:
[tex]\[ t = \frac{12.5 - 13.8}{1.639} \approx -0.795 \][/tex]
### 3. Determining the Degrees of Freedom (df):
The degrees of freedom for the test statistic is computed using the following formula:
[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{ \frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1} } \][/tex]
Calculations for numerator and denominator:
[tex]\[ df_{numerator} = \left( \frac{3.9^2}{14} + \frac{5.2^2}{17} \right)^2 \approx 2.688^2 \approx 7.224 \][/tex]
[tex]\[ df_{denominator} = \frac{\left( \frac{3.9^2}{14} \right)^2}{13} + \frac{\left( \frac{5.2^2}{17} \right)^2}{16} \approx \frac{0.0773^2}{13} + \frac{0.0941^2}{16} \approx \frac{0.005975}{13} + \frac{0.008859}{16} \approx 0.00046 + 0.00055 \approx 0.00101 \][/tex]
Therefore:
[tex]\[ df = \frac{7.224}{0.00101} \approx 28.79 \][/tex]
### 4. Finding the Critical Value:
Using the t-distribution table or a computational tool, we find the critical value for a one-tailed test at [tex]\( \alpha = 0.05 \)[/tex] with approximately 29 degrees of freedom. The critical value [tex]\( t_{critical} \)[/tex] is approximately:
[tex]\[ t_{critical} \approx -1.699 \][/tex]
### 5. Conclusion:
Comparing the test statistic to the critical value:
[tex]\[ t \approx -0.795 \quad \text{vs.} \quad t_{critical} \approx -1.699 \][/tex]
Since [tex]\( -0.795 \)[/tex] is greater than [tex]\( -1.699 \)[/tex]:
- We fail to reject the null hypothesis.
### Final Conclusion:
Based on the evidence from the sample, there is not enough statistical evidence at the 0.05 significance level to support the claim that women watch significantly less television than men. Thus, we conclude that there is no significant difference in television watching times between women and men.
### 1. Stating the Hypotheses:
We are given a significance level of [tex]\( \alpha = 0.05 \)[/tex]. The hypotheses for this problem are:
- Null Hypothesis [tex]\( H_0 \)[/tex]: The mean amount of time spent watching television by women is greater than or equal to the mean amount of time spent watching television by men.
[tex]\[ H_0: \mu_1 \geq \mu_2 \][/tex]
- Alternative Hypothesis [tex]\( H_a \)[/tex]: The mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.
[tex]\[ H_a: \mu_1 < \mu_2 \][/tex]
### 2. Calculating the Test Statistic:
We need to calculate the test statistic for the difference in means, assuming unequal variances. The formula for the test statistic [tex]\( t \)[/tex] is:
[tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} } \][/tex]
Given the summary statistics:
[tex]\[ \bar{x}_1 = 12.5 \, \text{hr}, \quad \bar{x}_2 = 13.8 \, \text{hr} \][/tex]
[tex]\[ s_1 = 3.9 \, \text{hr}, \quad s_2 = 5.2 \, \text{hr} \][/tex]
[tex]\[ n_1 = 14, \quad n_2 = 17 \][/tex]
First, we calculate the standard error (SE):
[tex]\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3.9^2}{14} + \frac{5.2^2}{17}} \approx 1.639 \][/tex]
Now the test statistic [tex]\( t \)[/tex] is:
[tex]\[ t = \frac{12.5 - 13.8}{1.639} \approx -0.795 \][/tex]
### 3. Determining the Degrees of Freedom (df):
The degrees of freedom for the test statistic is computed using the following formula:
[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{ \frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1} } \][/tex]
Calculations for numerator and denominator:
[tex]\[ df_{numerator} = \left( \frac{3.9^2}{14} + \frac{5.2^2}{17} \right)^2 \approx 2.688^2 \approx 7.224 \][/tex]
[tex]\[ df_{denominator} = \frac{\left( \frac{3.9^2}{14} \right)^2}{13} + \frac{\left( \frac{5.2^2}{17} \right)^2}{16} \approx \frac{0.0773^2}{13} + \frac{0.0941^2}{16} \approx \frac{0.005975}{13} + \frac{0.008859}{16} \approx 0.00046 + 0.00055 \approx 0.00101 \][/tex]
Therefore:
[tex]\[ df = \frac{7.224}{0.00101} \approx 28.79 \][/tex]
### 4. Finding the Critical Value:
Using the t-distribution table or a computational tool, we find the critical value for a one-tailed test at [tex]\( \alpha = 0.05 \)[/tex] with approximately 29 degrees of freedom. The critical value [tex]\( t_{critical} \)[/tex] is approximately:
[tex]\[ t_{critical} \approx -1.699 \][/tex]
### 5. Conclusion:
Comparing the test statistic to the critical value:
[tex]\[ t \approx -0.795 \quad \text{vs.} \quad t_{critical} \approx -1.699 \][/tex]
Since [tex]\( -0.795 \)[/tex] is greater than [tex]\( -1.699 \)[/tex]:
- We fail to reject the null hypothesis.
### Final Conclusion:
Based on the evidence from the sample, there is not enough statistical evidence at the 0.05 significance level to support the claim that women watch significantly less television than men. Thus, we conclude that there is no significant difference in television watching times between women and men.
