An online site presented this question, 'Would the recent norovirus outbreak deter you from taking a cruise?' Among 34,327 people who responded, 64% answered 'yes'. Use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond 'yes' to that question. Does the
To construct a confidence interval estimate for the proportion of the population who would respond 'yes' to the question, we can use the following formula:
Confidence Interval = Sample Proportion ± Margin of Error
To calculate the sample proportion, we divide the number of people who answered 'yes' by the total number of respondents:
Sample Proportion = Number of 'yes' responses / Total number of respondents
= (0.64 * 34,327) / 34,327
= 0.64
Since we want to construct a 90% confidence interval, we need to find the corresponding Z-score for this confidence level. The Z-score for a 90% confidence level is approximately 1.645.
Next, we calculate the margin of error using the following formula:
Margin of Error = Z-score * √((Sample Proportion * (1 - Sample Proportion)) / Sample Size)
In this case, the sample size is 34,327.
Margin of Error = 1.645 * √((0.64 * (1 - 0.64)) / 34,327)
≈ 0.008
Finally, we can construct the confidence interval by adding and subtracting the margin of error from the sample proportion:
Confidence Interval = Sample Proportion ± Margin of Error
= 0.64 ± 0.008
Therefore, the 90% confidence interval estimate for the proportion of the population who would respond 'yes' to the question is (0.632, 0.648). This means we are 90% confident that the true proportion of people who would respond 'yes' falls within this range.
To determine whether the recent norovirus outbreak would deter people from taking a cruise, we examine whether the confidence interval includes the null hypothesis value 0.5. In this case, since the confidence interval (0.632, 0.648) does not include 0.5, it suggests that the majority of respondents would be deterred from taking a cruise due to the recent norovirus outbreak.
To construct a confidence interval for the proportion of the population who would respond 'yes' to the question, we can use the formula:
CI = p̂ ± z*sqrt((p̂(1 - p̂))/n)
Where:
CI = Confidence Interval
p̂ = Sample proportion
z = Z-score corresponding to the desired confidence level
n = Sample size
In this case, the sample proportion (p̂) is 64% or 0.64, the sample size (n) is 34,327, and we want to construct a 90% confidence interval.
First, we need to calculate the Z-score for a 90% confidence level. The Z-score can be obtained from a standard normal distribution table or a statistical calculator. For a 90% confidence level, the Z-score is approximately 1.645.
Now we can substitute the values into the formula:
CI = 0.64 ± 1.645 * sqrt((0.64 * (1 - 0.64))/34,327)
Calculating the square root and simplifying:
CI ≈ 0.64 ± 1.645 * sqrt(0.0002222)
CI ≈ 0.64 ± 1.645 * 0.0149
CI ≈ 0.64 ± 0.0245
The 90% confidence interval is approximately (0.615, 0.665). This means that we can be 90% confident that the true proportion of people who would respond 'yes' to the question lies between 0.615 and 0.665.
To answer the second part of your question, we need to compare the confidence interval to a specific value. Since the confidence interval does not include 50%, it suggests that the proportion of people who would respond 'yes' is different from 50% at a 90% confidence level. It implies that a majority of people would be deterred by the recent norovirus outbreak.