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Answer: The 95% confidence interval will be wider.
Step-by-step explanation:
Confidence interval for population proportion is written as
Sample proportion ± margin of error
margin of error = z score × √pq/n
The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%
This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval. Therefore,
The 95% confidence interval will be wider.
The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
Given that,
If construct a 90% confidence interval for the population proportion,
And a 95% confidence interval for the population proportion,
Tim: 'The 90% confidence interval will be wider.'
Trevor: 'The 95% confidence interval will be wider.
Tracy: 'There is not enough information to tell.
Either interval could be wider.'
We have to determine,
Which of these intervals will be wider.
According to the question,
Three students provide their answers as follows:
Tim: 'The 90% confidence interval will be wider.
'Trevor: 'The 95% confidence interval will be wider.
'Tracy: 'There is not enough information to tell. Either interval could be wider.'
Therefore, Confidence interval for population proportion is written as
Sample proportion ± margin of error
Margin of error [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]
The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.
Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval.
Therefore, The 95% confidence interval will be wider.
Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
For more information about Probability click the link given below.
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