Answer :
Sure, let's go through the steps to answer this question in detail.
### Step 1: Define [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex]
- [tex]\( u_1 \)[/tex] is the population mean rating for males.
- [tex]\( u_2 \)[/tex] is the population mean rating for females.
These are the true average ratings for males and females, respectively, that we want to compare.
### Step 2: State the Hypotheses
- The null hypothesis ([tex]\( H_0 \)[/tex]) states that there is no difference in the mean ratings between males and females:
[tex]\[ H_0: u_1 = u_2 \][/tex]
- The alternative hypothesis ([tex]\( H_1 \)[/tex]) states that there is a difference in the mean ratings between males and females:
[tex]\[ H_1: u_1 \neq u_2 \][/tex]
### Step 3: Determine the Rejection Rule (Region)
The significance level ([tex]\( \alpha \)[/tex]) is given as 0.05. For a two-tailed test, we need to find the critical value of [tex]\( t \)[/tex] from the t-distribution that corresponds to this significance level.
- The rejection region will be determined by the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) such that the probability in each tail is [tex]\( \alpha/2 \)[/tex].
- The decision rule is: reject [tex]\( H_0 \)[/tex] if [tex]\( |t_{\text{statistic}}| > t_{\text{critical}} \)[/tex].
### Step 4: Calculate the Test Statistic
To calculate the test statistic, we use the formula for the pooled standard deviation when the sample sizes and variances are different:
[tex]\[ \text{pooled\_std} = \sqrt{\left( \frac{{\text{std}_1}^2}{n_1} + \frac{{\text{std}_2}^2}{n_2} \right)} \][/tex]
Substituting in the values:
[tex]\[ \text{pooled\_std} = \sqrt{\left( \frac{6.73^2}{30} + \frac{6.94^2}{32} \right)} = 4.623 \][/tex]
Next, we calculate the test statistic:
[tex]\[ t_{\text{statistic}} = \frac{\text{mean}_1 - \text{mean}_2}{\text{pooled\_std}} \][/tex]
Substituting in the values:
[tex]\[ t_{\text{statistic}} = \frac{39.08 - 38.79}{4.623} = 0.167 \][/tex]
To determine the degrees of freedom (df) for the t-distribution, we use the following formula for unequal variances:
[tex]\[ df = \frac{\left( \frac{{\text{std}_1}^2}{n_1} + \frac{{\text{std}_2}^2}{n_2} \right)^2}{\left( \frac{{\text{std}_1}^2/n_1}^2 / (n_1-1) \right) + \left( \frac{{\text{std}_2}^2/n_2}^2 / (n_2-1) \right)} \][/tex]
The degrees of freedom (df) calculated using the above values is approximately 58. This value is then used to find the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) for a two-tailed test at [tex]\( \alpha = 0.05 \)[/tex]. The critical value [tex]\( t_{\text{critical}} \)[/tex] from the t-distribution table is approximately 2.000.
### Step 5: Make the Decision
With the calculated test statistic of [tex]\( t_{\text{statistic}} = 0.167 \)[/tex] and the critical value [tex]\( t_{\text{critical}} = 2.000 \)[/tex]:
- Because [tex]\( |0.167| < 2.000 \)[/tex], we fail to reject the null hypothesis.
### Step 6: Conclusion
Based on this analysis, we fail to reject the null hypothesis. Therefore, we conclude that there is no significant evidence to suggest that the mean ratings of service quality at five-star hotels in Jamaica are different between males and females.
### Step 1: Define [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex]
- [tex]\( u_1 \)[/tex] is the population mean rating for males.
- [tex]\( u_2 \)[/tex] is the population mean rating for females.
These are the true average ratings for males and females, respectively, that we want to compare.
### Step 2: State the Hypotheses
- The null hypothesis ([tex]\( H_0 \)[/tex]) states that there is no difference in the mean ratings between males and females:
[tex]\[ H_0: u_1 = u_2 \][/tex]
- The alternative hypothesis ([tex]\( H_1 \)[/tex]) states that there is a difference in the mean ratings between males and females:
[tex]\[ H_1: u_1 \neq u_2 \][/tex]
### Step 3: Determine the Rejection Rule (Region)
The significance level ([tex]\( \alpha \)[/tex]) is given as 0.05. For a two-tailed test, we need to find the critical value of [tex]\( t \)[/tex] from the t-distribution that corresponds to this significance level.
- The rejection region will be determined by the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) such that the probability in each tail is [tex]\( \alpha/2 \)[/tex].
- The decision rule is: reject [tex]\( H_0 \)[/tex] if [tex]\( |t_{\text{statistic}}| > t_{\text{critical}} \)[/tex].
### Step 4: Calculate the Test Statistic
To calculate the test statistic, we use the formula for the pooled standard deviation when the sample sizes and variances are different:
[tex]\[ \text{pooled\_std} = \sqrt{\left( \frac{{\text{std}_1}^2}{n_1} + \frac{{\text{std}_2}^2}{n_2} \right)} \][/tex]
Substituting in the values:
[tex]\[ \text{pooled\_std} = \sqrt{\left( \frac{6.73^2}{30} + \frac{6.94^2}{32} \right)} = 4.623 \][/tex]
Next, we calculate the test statistic:
[tex]\[ t_{\text{statistic}} = \frac{\text{mean}_1 - \text{mean}_2}{\text{pooled\_std}} \][/tex]
Substituting in the values:
[tex]\[ t_{\text{statistic}} = \frac{39.08 - 38.79}{4.623} = 0.167 \][/tex]
To determine the degrees of freedom (df) for the t-distribution, we use the following formula for unequal variances:
[tex]\[ df = \frac{\left( \frac{{\text{std}_1}^2}{n_1} + \frac{{\text{std}_2}^2}{n_2} \right)^2}{\left( \frac{{\text{std}_1}^2/n_1}^2 / (n_1-1) \right) + \left( \frac{{\text{std}_2}^2/n_2}^2 / (n_2-1) \right)} \][/tex]
The degrees of freedom (df) calculated using the above values is approximately 58. This value is then used to find the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) for a two-tailed test at [tex]\( \alpha = 0.05 \)[/tex]. The critical value [tex]\( t_{\text{critical}} \)[/tex] from the t-distribution table is approximately 2.000.
### Step 5: Make the Decision
With the calculated test statistic of [tex]\( t_{\text{statistic}} = 0.167 \)[/tex] and the critical value [tex]\( t_{\text{critical}} = 2.000 \)[/tex]:
- Because [tex]\( |0.167| < 2.000 \)[/tex], we fail to reject the null hypothesis.
### Step 6: Conclusion
Based on this analysis, we fail to reject the null hypothesis. Therefore, we conclude that there is no significant evidence to suggest that the mean ratings of service quality at five-star hotels in Jamaica are different between males and females.
