Answer :
Let's tackle the question step-by-step.
(a) Checking the Conditions for a Confidence Interval:
To calculate a confidence interval for a proportion, certain conditions must be satisfied:
1. Random Sample: The question states that a simple random sample (SRS) of 172 undergraduates was used, so this condition is satisfied.
2. Large Counts Condition: This requires that both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are at least 10, where [tex]\( \hat{p} \)[/tex] is the sample proportion.
- The number of students surveyed, [tex]\( n \)[/tex], is 172.
- The number of students who answered "Yes" is 19. So the sample proportion [tex]\( \hat{p} \)[/tex] is 19/172.
Let's calculate:
[tex]\[
n \hat{p} = 172 \times \frac{19}{172} = 19
\][/tex]
[tex]\[
n(1 - \hat{p}) = 172 \times \left(1 - \frac{19}{172}\right) = 153
\][/tex]
Both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are greater than 10, so the large counts condition is satisfied.
(b) Calculating the 99% Confidence Interval:
To calculate the confidence interval, we need:
- The sample proportion [tex]\( \hat{p} = \frac{19}{172} \)[/tex].
- The critical value [tex]\( z^* \)[/tex] for a 99% confidence level, which is approximately 2.576.
- The standard error of the proportion:
[tex]\[
\text{Standard Error} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\][/tex]
Let's calculate the margin of error:
[tex]\[
\text{Margin of Error} = z^* \times \text{Standard Error}
\][/tex]
Finally, the confidence interval is:
[tex]\[
(\hat{p} - \text{Margin of Error}, \hat{p} + \text{Margin of Error})
\][/tex]
Substituting the values, we find:
[tex]\[
(0.049, 0.172)
\][/tex]
(c) Interpreting the Interval:
The 99% confidence interval (0.049 to 0.172) means that we are 99% confident that the true proportion of all undergraduate students at this university who would answer "Yes" to reporting cheating lies between 4.9% and 17.2%. This interval provides a range of plausible values for the proportion based on the sample data.
(a) Checking the Conditions for a Confidence Interval:
To calculate a confidence interval for a proportion, certain conditions must be satisfied:
1. Random Sample: The question states that a simple random sample (SRS) of 172 undergraduates was used, so this condition is satisfied.
2. Large Counts Condition: This requires that both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are at least 10, where [tex]\( \hat{p} \)[/tex] is the sample proportion.
- The number of students surveyed, [tex]\( n \)[/tex], is 172.
- The number of students who answered "Yes" is 19. So the sample proportion [tex]\( \hat{p} \)[/tex] is 19/172.
Let's calculate:
[tex]\[
n \hat{p} = 172 \times \frac{19}{172} = 19
\][/tex]
[tex]\[
n(1 - \hat{p}) = 172 \times \left(1 - \frac{19}{172}\right) = 153
\][/tex]
Both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are greater than 10, so the large counts condition is satisfied.
(b) Calculating the 99% Confidence Interval:
To calculate the confidence interval, we need:
- The sample proportion [tex]\( \hat{p} = \frac{19}{172} \)[/tex].
- The critical value [tex]\( z^* \)[/tex] for a 99% confidence level, which is approximately 2.576.
- The standard error of the proportion:
[tex]\[
\text{Standard Error} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\][/tex]
Let's calculate the margin of error:
[tex]\[
\text{Margin of Error} = z^* \times \text{Standard Error}
\][/tex]
Finally, the confidence interval is:
[tex]\[
(\hat{p} - \text{Margin of Error}, \hat{p} + \text{Margin of Error})
\][/tex]
Substituting the values, we find:
[tex]\[
(0.049, 0.172)
\][/tex]
(c) Interpreting the Interval:
The 99% confidence interval (0.049 to 0.172) means that we are 99% confident that the true proportion of all undergraduate students at this university who would answer "Yes" to reporting cheating lies between 4.9% and 17.2%. This interval provides a range of plausible values for the proportion based on the sample data.
