To solve this problem, we need to calculate a 95% confidence interval for the population proportion $p$ of adults in city X who own a mobile phone.

Step-by-Step Solution:

1. Define the Sample Proportion:

   • Given:
     • Total sample size $n = 426$
     • Number of people who own a phone $x = 153$
   • Sample proportion $\hat{p}$ is calculated as: $\hat{p} = \frac{x}{n} = \frac{153}{426} \approx 0.3592$

2. Determine the Confidence Level and Critical Value:

   • For a 95% confidence level, the critical value $z$ (for a standard normal distribution) is approximately 1.96.

3. Calculate the Standard Error (SE):

   • Standard error for the proportion is given by: $SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$
   • Substituting the values: $SE = \sqrt{\frac{0.3592 \times (1 - 0.3592)}{426}} \approx 0.0233$

4. Calculate the Margin of Error (ME):

   • The margin of error is $ME = z \times SE$: $ME = 1.96 \times 0.0233 \approx 0.0457$

5. Find the Confidence Interval:

   • The 95% confidence interval is given by: $\hat{p} - ME < p < \hat{p} + ME$
   • Substituting the values: $0.3592 - 0.0457 < p < 0.3592 + 0.0457$ $0.3135 < p < 0.4049$

Answer:

The 95% confidence interval for $p$ is approximately: $0.31 < p < 0.40$

Would you like more details or have any questions?

Related Questions:

1. How would the confidence interval change if the confidence level was increased to 99%?
2. What is the effect of a larger sample size on the width of the confidence interval?
3. How would we calculate the confidence interval if the sample proportion was unknown?
4. How does changing the sample size affect the standard error of the proportion?
5. What assumptions are necessary for constructing a confidence interval in this scenario?

Tip:

For a larger sample, the confidence interval becomes narrower, meaning we estimate the true proportion more precisely.