Answer :
Sure, let's go through this step-by-step.
### Step-by-Step Solution
#### Part (a): Construct and interpret a 99% confidence interval
Given:
- Sample size ([tex]\(n\)[/tex]) = 508
- Sample proportion ([tex]\(\hat{p}\)[/tex]) = 0.31
- Confidence level = 99%
To construct a 99% confidence interval for the population proportion, we need to:
1. Calculate the standard error ([tex]\(SE\)[/tex])
2. Determine the z-critical value ([tex]\(z^\)[/tex])
3. Compute the margin of error ([tex]\(ME\)[/tex])
4. Construct the confidence interval
1. Calculate the standard error (SE):
[tex]\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
Plugging in the values:
[tex]\[ SE = \sqrt{\frac{0.31 \times (1 - 0.31)}{508}} = \sqrt{\frac{0.31 \times 0.69}{508}} \approx 0.0209 \][/tex]
2. Determine the z-critical value ([tex]\(z^\)[/tex]):
For a 99% confidence level, the [tex]\(z^\)[/tex] value is found in standard normal distribution tables (or using statistical tools). It is approximately 2.576.
3. Compute the margin of error (ME):
[tex]\[ ME = z^ \times SE \][/tex]
[tex]\[ ME = 2.576 \times 0.0209 \approx 0.0539 \][/tex]
4. Construct the confidence interval:
The confidence interval is given by:
[tex]\[ \hat{p} \pm ME \][/tex]
[tex]\[ 0.31 \pm 0.0539 \][/tex]
So, the 99% confidence interval is:
[tex]\[ (0.2561, 0.3639) \][/tex]
Interpretation:
We are 99% confident that the true proportion of all U.S. residents who believe life ever existed on Mars is between 25.61% and 36.39%.
#### Part (b): Is it plausible that more than 35% of U.S. residents think life ever existed on Mars?
To determine if it's plausible that more than 35% of U.S. residents think life ever existed on Mars, we need to check if 0.35 is within the confidence interval we just calculated.
The confidence interval is (0.2561, 0.3639). Since 0.35 is within this interval, it is plausible that more than 35% of U.S. residents could believe that life ever existed on Mars. However, since 0.35 is close to the upper limit of the interval, we would say it is plausible but not highly likely.
#### Part (c): Calculating the required sample size to achieve a margin of error of at most 4%
To achieve a new desired margin of error ([tex]\(ME_{desired}\)[/tex]) of 0.04, we can use the formula:
[tex]\[ ME_{desired} = z^ \times SE_{desired} \][/tex]
First, solve for the new standard error:
[tex]\[ SE_{desired} = \frac{ME_{desired}}{z^} \][/tex]
Given [tex]\( z^* = 2.576 \)[/tex] and [tex]\( ME_{desired} = 0.04 \)[/tex]:
[tex]\[ SE_{desired} = \frac{0.04}{2.576} \approx 0.0155 \][/tex]
Next, we use the standard error formula to find the new required sample size ([tex]\( n_{new} \)[/tex]):
[tex]\[ SE_{desired} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n_{new}}} \][/tex]
Plugging in [tex]\( \hat{p} = 0.31 \)[/tex]:
[tex]\[ 0.0155 = \sqrt{\frac{0.31 \times 0.69}{n_{new}}} \][/tex]
Square both sides to solve for [tex]\( n_{new} \)[/tex]:
[tex]\[ 0.0155^2 = \frac{0.31 \times 0.69}{n_{new}} \][/tex]
[tex]\[ 0.00024025 = \frac{0.2139}{n_{new}} \][/tex]
[tex]\[ n_{new} = \frac{0.2139}{0.00024025} \approx 890 \][/tex]
The current sample size is 508, so the additional number of U.S. residents needed:
[tex]\[ \text{Additional sample size} = n_{new} - 508 \][/tex]
[tex]\[ \text{Additional sample size} = 890 - 508 = 382 \][/tex]
Therefore, to achieve a margin of error of at most 4%, the polltakers need to randomly select about 382 additional U.S. residents.
### Step-by-Step Solution
#### Part (a): Construct and interpret a 99% confidence interval
Given:
- Sample size ([tex]\(n\)[/tex]) = 508
- Sample proportion ([tex]\(\hat{p}\)[/tex]) = 0.31
- Confidence level = 99%
To construct a 99% confidence interval for the population proportion, we need to:
1. Calculate the standard error ([tex]\(SE\)[/tex])
2. Determine the z-critical value ([tex]\(z^\)[/tex])
3. Compute the margin of error ([tex]\(ME\)[/tex])
4. Construct the confidence interval
1. Calculate the standard error (SE):
[tex]\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
Plugging in the values:
[tex]\[ SE = \sqrt{\frac{0.31 \times (1 - 0.31)}{508}} = \sqrt{\frac{0.31 \times 0.69}{508}} \approx 0.0209 \][/tex]
2. Determine the z-critical value ([tex]\(z^\)[/tex]):
For a 99% confidence level, the [tex]\(z^\)[/tex] value is found in standard normal distribution tables (or using statistical tools). It is approximately 2.576.
3. Compute the margin of error (ME):
[tex]\[ ME = z^ \times SE \][/tex]
[tex]\[ ME = 2.576 \times 0.0209 \approx 0.0539 \][/tex]
4. Construct the confidence interval:
The confidence interval is given by:
[tex]\[ \hat{p} \pm ME \][/tex]
[tex]\[ 0.31 \pm 0.0539 \][/tex]
So, the 99% confidence interval is:
[tex]\[ (0.2561, 0.3639) \][/tex]
Interpretation:
We are 99% confident that the true proportion of all U.S. residents who believe life ever existed on Mars is between 25.61% and 36.39%.
#### Part (b): Is it plausible that more than 35% of U.S. residents think life ever existed on Mars?
To determine if it's plausible that more than 35% of U.S. residents think life ever existed on Mars, we need to check if 0.35 is within the confidence interval we just calculated.
The confidence interval is (0.2561, 0.3639). Since 0.35 is within this interval, it is plausible that more than 35% of U.S. residents could believe that life ever existed on Mars. However, since 0.35 is close to the upper limit of the interval, we would say it is plausible but not highly likely.
#### Part (c): Calculating the required sample size to achieve a margin of error of at most 4%
To achieve a new desired margin of error ([tex]\(ME_{desired}\)[/tex]) of 0.04, we can use the formula:
[tex]\[ ME_{desired} = z^ \times SE_{desired} \][/tex]
First, solve for the new standard error:
[tex]\[ SE_{desired} = \frac{ME_{desired}}{z^} \][/tex]
Given [tex]\( z^* = 2.576 \)[/tex] and [tex]\( ME_{desired} = 0.04 \)[/tex]:
[tex]\[ SE_{desired} = \frac{0.04}{2.576} \approx 0.0155 \][/tex]
Next, we use the standard error formula to find the new required sample size ([tex]\( n_{new} \)[/tex]):
[tex]\[ SE_{desired} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n_{new}}} \][/tex]
Plugging in [tex]\( \hat{p} = 0.31 \)[/tex]:
[tex]\[ 0.0155 = \sqrt{\frac{0.31 \times 0.69}{n_{new}}} \][/tex]
Square both sides to solve for [tex]\( n_{new} \)[/tex]:
[tex]\[ 0.0155^2 = \frac{0.31 \times 0.69}{n_{new}} \][/tex]
[tex]\[ 0.00024025 = \frac{0.2139}{n_{new}} \][/tex]
[tex]\[ n_{new} = \frac{0.2139}{0.00024025} \approx 890 \][/tex]
The current sample size is 508, so the additional number of U.S. residents needed:
[tex]\[ \text{Additional sample size} = n_{new} - 508 \][/tex]
[tex]\[ \text{Additional sample size} = 890 - 508 = 382 \][/tex]
Therefore, to achieve a margin of error of at most 4%, the polltakers need to randomly select about 382 additional U.S. residents.
