A survey of 122 randomly selected students at one college showed that only 90 checked their campus email account on a regular basis. Construct a 90% confidence interval for the percentage of students at the college who do not check their email account on a regular basis. Round to one decimal place.

To construct a 90% confidence interval for the percentage of students who **do not** check their email on a regular basis, let's proceed with the following steps:

1. **Calculate the sample proportion of students who do not check their email regularly**:
   \[
   \hat{p} = \frac{\text{Number of students who do not check email}}{\text{Total students surveyed}} = \frac{122 - 90}{122} = \frac{32}{122} \approx 0.2623
   \]

2. **Determine the complement of the confidence level**:  
   Since we want a 90% confidence level, the complement is \( 1 - 0.90 = 0.10 \). Dividing by 2 for the two tails of the normal distribution gives \( \alpha/2 = 0.05 \).

3. **Find the critical z-value for a 90% confidence level**:
   For a 90% confidence interval, the critical z-value (\( z_{0.05} \)) is approximately 1.645.

4. **Calculate the standard error**:
   The standard error \( SE \) for a proportion is given by:
   \[
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.2623 \times (1 - 0.2623)}{122}} \approx 0.0407
   \]

5. **Construct the confidence interval**:
   The confidence interval is given by:
   \[
   \hat{p} \pm z \times SE
   \]
   Substituting in our values:
   \[
   0.2623 \pm 1.645 \times 0.0407 \approx 0.2623 \pm 0.0669
   \]
   This gives an interval of approximately \( (0.1954, 0.3292) \).

6. **Convert to a percentage and round**:
   \[
   0.1954 \times 100 \approx 19.5\% \quad \text{and} \quad 0.3292 \times 100 \approx 32.9\%
   \]
   So, the 90% confidence interval for the percentage of students who do not check their email regularly is **(19.5%, 32.9%)**.

Would you like more details on any of these steps, or have further questions?

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Here are five related questions to deepen understanding:

1. How would the confidence interval change with a 95% confidence level?
2. What effect does increasing the sample size have on the width of the confidence interval?
3. How would the confidence interval change if the proportion of students not checking email was higher?
4. Why is it important to check if the sample is random when constructing confidence intervals?
5. How would the interpretation differ if this was a one-sided confidence interval?

**Tip**: Increasing the sample size decreases the standard error, making the confidence interval narrower and more precise.

Construct a 90% confidence interval for the percentage of students at a college who do not regularly check their campus email. This problem involves key statistical concepts such as sample proportions and confidence intervals, with detailed calculations for a precise answer. Suitable for high school students in Grades 10-12.

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