Answer :
To draw a conclusion for the hypothesis test, we follow a systematic approach. Here are the steps broken down:
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The number of customers served each day is uniform across the week (i.e., the same number of customers are served each day).
- Alternative Hypothesis ([tex]\(H_A\)[/tex]): The number of customers served each day is not uniform (i.e., the number of customers served each day differs).
2. Significance Level ([tex]\(\alpha\)[/tex]):
- The significance level given is [tex]\( \alpha = 0.05 \)[/tex] (5%).
3. Test Statistic:
- A Chi-Square goodness-of-fit test is appropriate to compare the observed distribution of customers with the uniform distribution.
4. Determine the p-value:
- The problem provides the [tex]\( p \)[/tex]-value, which is [tex]\( p = 0.01 \)[/tex].
5. Compare the p-value with the significance level:
- Compare [tex]\( p = 0.01 \)[/tex] with [tex]\( \alpha = 0.05 \)[/tex].
Since [tex]\( p \)[/tex] (0.01) is less than [tex]\( \alpha \)[/tex] (0.05), we have enough evidence to reject the null hypothesis.
6. Conclusion:
- Because the p-value is less than the significance level, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the number of customers served each day is not the same.
In summary, based on the Chi-Square goodness-of-fit test with a [tex]\( p \)[/tex]-value of 0.01 and a significance level of 0.05, we reject the null hypothesis. This leads us to conclude that there is sufficient evidence to suggest that the number of customers served each day of the week is not uniform.
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The number of customers served each day is uniform across the week (i.e., the same number of customers are served each day).
- Alternative Hypothesis ([tex]\(H_A\)[/tex]): The number of customers served each day is not uniform (i.e., the number of customers served each day differs).
2. Significance Level ([tex]\(\alpha\)[/tex]):
- The significance level given is [tex]\( \alpha = 0.05 \)[/tex] (5%).
3. Test Statistic:
- A Chi-Square goodness-of-fit test is appropriate to compare the observed distribution of customers with the uniform distribution.
4. Determine the p-value:
- The problem provides the [tex]\( p \)[/tex]-value, which is [tex]\( p = 0.01 \)[/tex].
5. Compare the p-value with the significance level:
- Compare [tex]\( p = 0.01 \)[/tex] with [tex]\( \alpha = 0.05 \)[/tex].
Since [tex]\( p \)[/tex] (0.01) is less than [tex]\( \alpha \)[/tex] (0.05), we have enough evidence to reject the null hypothesis.
6. Conclusion:
- Because the p-value is less than the significance level, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the number of customers served each day is not the same.
In summary, based on the Chi-Square goodness-of-fit test with a [tex]\( p \)[/tex]-value of 0.01 and a significance level of 0.05, we reject the null hypothesis. This leads us to conclude that there is sufficient evidence to suggest that the number of customers served each day of the week is not uniform.