Answer :
To determine if the data provide convincing evidence that the proportion of rolls resulting in 6 spots up is different from [tex]\(\frac{1}{6}\)[/tex], we can conduct a hypothesis test for a proportion. Let's break this down step-by-step:
1. Define Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The proportion of 6 spots up ([tex]\(p\)[/tex]) is [tex]\(\frac{1}{6}\)[/tex].
[tex]\[
H_0: p = \frac{1}{6}
\][/tex]
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The proportion of 6 spots up is different from [tex]\(\frac{1}{6}\)[/tex].
[tex]\[
H_a: p \neq \frac{1}{6}
\][/tex]
2. Check Conditions for Inference:
- Random Condition: We should have a random sample. Here, we assume the sample is random.
- 10% Condition: The sample size [tex]\(n\)[/tex] should be less than 10% of the population. Assuming the population of cubes is greater than 1500, this condition is not met since [tex]\(150 \not< 0.10 \times 1500\)[/tex].
- Large Counts Condition: We need at least 10 expected successes and 10 expected failures.
- Expected successes: [tex]\(n \times \frac{1}{6} = 150 \times \frac{1}{6} = 25\)[/tex]
- Expected failures: [tex]\(n \times (1 - \frac{1}{6}) = 150 \times \frac{5}{6} = 125\)[/tex]
- Both are greater than 10, so this condition is met.
3. Calculate Sample Proportion:
The sample proportion ([tex]\(p_{\text{hat}}\)[/tex]) is calculated as:
[tex]\[
p_{\text{hat}} = \frac{\text{Number of successes}}{n} = \frac{35}{150} \approx 0.2333
\][/tex]
4. Calculate Standard Error:
The standard error (SE) is calculated using the formula for the standard error of a proportion:
[tex]\[
\text{SE} = \sqrt{\frac{p_{\text{null}}(1 - p_{\text{null}})}{n}} = \sqrt{\frac{\frac{1}{6} \times \frac{5}{6}}{150}} \approx 0.0304
\][/tex]
5. Calculate Test Statistic (z-score):
The z-score is calculated as:
[tex]\[
z = \frac{p_{\text{hat}} - p_{\text{null}}}{\text{SE}} = \frac{0.2333 - \frac{1}{6}}{0.0304} \approx 2.19
\][/tex]
6. Calculate p-value:
Since this is a two-tailed test, the p-value is found using the standard normal distribution:
[tex]\[
p\text{-value} = 2 \times \text{P}(Z > |2.19|) \approx 0.0285
\][/tex]
7. Conclusion:
To conclude, compare the p-value with the significance level [tex]\(\alpha = 0.01\)[/tex]:
- Since [tex]\(0.0285 > 0.01\)[/tex], we do not have enough evidence to reject the null hypothesis at the 0.01 significance level.
Therefore, at the [tex]\(\alpha = 0.01\)[/tex] significance level, the data do not provide convincing evidence that the proportion of rolls resulting in 6 spots up is different from [tex]\(\frac{1}{6}\)[/tex].
8. Summary of Conditions:
- Random Condition: Met (assumed).
- 10% Condition: Not met.
- Large Counts Condition: Met (expected successes and failures are sufficient).
This means the inference might be affected because the 10% condition isn't satisfied, but the other conditions were fine, so take caution in interpretation.
1. Define Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The proportion of 6 spots up ([tex]\(p\)[/tex]) is [tex]\(\frac{1}{6}\)[/tex].
[tex]\[
H_0: p = \frac{1}{6}
\][/tex]
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The proportion of 6 spots up is different from [tex]\(\frac{1}{6}\)[/tex].
[tex]\[
H_a: p \neq \frac{1}{6}
\][/tex]
2. Check Conditions for Inference:
- Random Condition: We should have a random sample. Here, we assume the sample is random.
- 10% Condition: The sample size [tex]\(n\)[/tex] should be less than 10% of the population. Assuming the population of cubes is greater than 1500, this condition is not met since [tex]\(150 \not< 0.10 \times 1500\)[/tex].
- Large Counts Condition: We need at least 10 expected successes and 10 expected failures.
- Expected successes: [tex]\(n \times \frac{1}{6} = 150 \times \frac{1}{6} = 25\)[/tex]
- Expected failures: [tex]\(n \times (1 - \frac{1}{6}) = 150 \times \frac{5}{6} = 125\)[/tex]
- Both are greater than 10, so this condition is met.
3. Calculate Sample Proportion:
The sample proportion ([tex]\(p_{\text{hat}}\)[/tex]) is calculated as:
[tex]\[
p_{\text{hat}} = \frac{\text{Number of successes}}{n} = \frac{35}{150} \approx 0.2333
\][/tex]
4. Calculate Standard Error:
The standard error (SE) is calculated using the formula for the standard error of a proportion:
[tex]\[
\text{SE} = \sqrt{\frac{p_{\text{null}}(1 - p_{\text{null}})}{n}} = \sqrt{\frac{\frac{1}{6} \times \frac{5}{6}}{150}} \approx 0.0304
\][/tex]
5. Calculate Test Statistic (z-score):
The z-score is calculated as:
[tex]\[
z = \frac{p_{\text{hat}} - p_{\text{null}}}{\text{SE}} = \frac{0.2333 - \frac{1}{6}}{0.0304} \approx 2.19
\][/tex]
6. Calculate p-value:
Since this is a two-tailed test, the p-value is found using the standard normal distribution:
[tex]\[
p\text{-value} = 2 \times \text{P}(Z > |2.19|) \approx 0.0285
\][/tex]
7. Conclusion:
To conclude, compare the p-value with the significance level [tex]\(\alpha = 0.01\)[/tex]:
- Since [tex]\(0.0285 > 0.01\)[/tex], we do not have enough evidence to reject the null hypothesis at the 0.01 significance level.
Therefore, at the [tex]\(\alpha = 0.01\)[/tex] significance level, the data do not provide convincing evidence that the proportion of rolls resulting in 6 spots up is different from [tex]\(\frac{1}{6}\)[/tex].
8. Summary of Conditions:
- Random Condition: Met (assumed).
- 10% Condition: Not met.
- Large Counts Condition: Met (expected successes and failures are sufficient).
This means the inference might be affected because the 10% condition isn't satisfied, but the other conditions were fine, so take caution in interpretation.
