Out of 100 people sampled, 70 received flu vaccinations this year. Based on this, construct a $95 \%$ confidence interval for the true population proportion of people who received flu vaccinations this year.
Give your answers as decimals, to three places
$< p< $

Step 1 ：The problem is asking for a confidence interval for a population proportion. The formula for a confidence interval is given by: \[\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired level of confidence, and \(n\) is the sample size.

Step 2 ：In this case, \(\hat{p} = \frac{70}{100} = 0.7\), \(Z = 1.96\) for a 95% confidence interval, and \(n = 100\).

Step 3 ：We can plug these values into the formula to find the confidence interval.

Step 4 ：Calculate the standard error (SE) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). Substituting the given values, we get \(SE = 0.045825756949558406\).

Step 5 ：Calculate the lower and upper bounds of the confidence interval using the formula \(\hat{p} \pm Z \times SE\). Substituting the given values, we get \(CI_{lower} = 0.6101815163788655\) and \(CI_{upper} = 0.7898184836211344\).

Step 6 ：Final Answer: The 95% confidence interval for the true population proportion of people who received flu vaccinations this year is \(\boxed{(0.610, 0.790)}\).