Answer :
Sure, let's tackle this problem step-by-step.
### (a) Show that the conditions for calculating a confidence interval for a proportion are satisfied.
To calculate a confidence interval for a proportion, we need to ensure that certain conditions are met:
1. The sample should be a Simple Random Sample (SRS) from the population.
2. The sample size should be large enough such that both [tex]\[ np \][/tex] and [tex]\[ n(1-p) \][/tex] are greater than 5, where [tex]\[ n \][/tex] is the sample size, and [tex]\[ p \][/tex] is the sample proportion ([tex]\[ p-hat \][/tex]).
Given:
- Sample size ([tex]\[ n \][/tex]) = 172
- Number of successes (students answered "Yes") ([tex]\[ x \][/tex]) = 19
First, let's calculate [tex]\[ p-hat \][/tex]:
[tex]\[ p-hat = \frac{x}{n} = \frac{19}{172} \approx 0.1105\][/tex]
Now, check the conditions:
- [tex]\[ n p-hat = 172 0.1105 \approx 19 \][/tex]
- [tex]\[ n (1 - p-hat) = 172 (1 - 0.1105) = 172 0.8895 \approx 153.1 \][/tex]
Both conditions, [tex]\[ n p-hat > 5 \][/tex] and [tex]\[ n (1 - p-hat) > 5 \][/tex], are satisfied because 19 > 5 and 153.1 > 5.
### (b) Calculate a 99% confidence interval for the proportion of all undergraduate students at this university who would answer "Yes."
To calculate the confidence interval:
1. Sample Proportion ([tex]\[ p-hat \][/tex]):
[tex]\[ p-hat = 0.1105 \][/tex]
2. Standard Error ([tex]\[ SE \][/tex]):
[tex]\[ SE = \sqrt{ \frac{ p-hat (1 - p-hat) }{ n } } = \sqrt{ \frac{ 0.1105 (1 - 0.1105) }{ 172 } } \approx \sqrt{ \frac{ 0.1105 \times 0.8895 }{ 172 } } \approx \sqrt{ \frac{ 0.0982 }{ 172 } } \approx \sqrt{0.000570} \approx 0.0239 \][/tex]
3. Z-score for a 99% confidence level (using standard normal distribution tables):
For a 99% confidence interval, [tex]\[ (1 - \alpha) = 0.99 \][/tex], so [tex]\[ \alpha/2 = 0.01/2 = 0.005 \][/tex]. The Z-score corresponding to 0.005 in the standard normal distribution is approximately 2.576.
4. Margin of Error ([tex]\[ ME \][/tex]):
[tex]\[ ME = Z SE = 2.576 * 0.0239 \approx 0.0616 \][/tex]
5. Confidence Interval (CI):
[tex]\[ CI_{lower} = p-hat - ME = 0.1105 - 0.0616 = 0.0489 \][/tex]
[tex]\[ CI_{upper} = p-hat + ME = 0.1105 + 0.0616 = 0.1721 \][/tex]
So, the 99% confidence interval is approximately (0.0489, 0.1721).
### (c) Interpret the interval from part (b).
The 99% confidence interval for the proportion of all undergraduate students at this university who would answer "Yes" to reporting cheating is approximately between 4.89% and 17.21%. This means that we are 99% confident that the true proportion of students who would report cheating falls within this interval. In other words, if we were to take many samples like this, 99% of the computed confidence intervals would contain the true proportion of students willing to report cheating.
### (a) Show that the conditions for calculating a confidence interval for a proportion are satisfied.
To calculate a confidence interval for a proportion, we need to ensure that certain conditions are met:
1. The sample should be a Simple Random Sample (SRS) from the population.
2. The sample size should be large enough such that both [tex]\[ np \][/tex] and [tex]\[ n(1-p) \][/tex] are greater than 5, where [tex]\[ n \][/tex] is the sample size, and [tex]\[ p \][/tex] is the sample proportion ([tex]\[ p-hat \][/tex]).
Given:
- Sample size ([tex]\[ n \][/tex]) = 172
- Number of successes (students answered "Yes") ([tex]\[ x \][/tex]) = 19
First, let's calculate [tex]\[ p-hat \][/tex]:
[tex]\[ p-hat = \frac{x}{n} = \frac{19}{172} \approx 0.1105\][/tex]
Now, check the conditions:
- [tex]\[ n p-hat = 172 0.1105 \approx 19 \][/tex]
- [tex]\[ n (1 - p-hat) = 172 (1 - 0.1105) = 172 0.8895 \approx 153.1 \][/tex]
Both conditions, [tex]\[ n p-hat > 5 \][/tex] and [tex]\[ n (1 - p-hat) > 5 \][/tex], are satisfied because 19 > 5 and 153.1 > 5.
### (b) Calculate a 99% confidence interval for the proportion of all undergraduate students at this university who would answer "Yes."
To calculate the confidence interval:
1. Sample Proportion ([tex]\[ p-hat \][/tex]):
[tex]\[ p-hat = 0.1105 \][/tex]
2. Standard Error ([tex]\[ SE \][/tex]):
[tex]\[ SE = \sqrt{ \frac{ p-hat (1 - p-hat) }{ n } } = \sqrt{ \frac{ 0.1105 (1 - 0.1105) }{ 172 } } \approx \sqrt{ \frac{ 0.1105 \times 0.8895 }{ 172 } } \approx \sqrt{ \frac{ 0.0982 }{ 172 } } \approx \sqrt{0.000570} \approx 0.0239 \][/tex]
3. Z-score for a 99% confidence level (using standard normal distribution tables):
For a 99% confidence interval, [tex]\[ (1 - \alpha) = 0.99 \][/tex], so [tex]\[ \alpha/2 = 0.01/2 = 0.005 \][/tex]. The Z-score corresponding to 0.005 in the standard normal distribution is approximately 2.576.
4. Margin of Error ([tex]\[ ME \][/tex]):
[tex]\[ ME = Z SE = 2.576 * 0.0239 \approx 0.0616 \][/tex]
5. Confidence Interval (CI):
[tex]\[ CI_{lower} = p-hat - ME = 0.1105 - 0.0616 = 0.0489 \][/tex]
[tex]\[ CI_{upper} = p-hat + ME = 0.1105 + 0.0616 = 0.1721 \][/tex]
So, the 99% confidence interval is approximately (0.0489, 0.1721).
### (c) Interpret the interval from part (b).
The 99% confidence interval for the proportion of all undergraduate students at this university who would answer "Yes" to reporting cheating is approximately between 4.89% and 17.21%. This means that we are 99% confident that the true proportion of students who would report cheating falls within this interval. In other words, if we were to take many samples like this, 99% of the computed confidence intervals would contain the true proportion of students willing to report cheating.
