1. **State the problem:** We want to test if the true proportion of students owning laptops differs from the university's claim of 70%. 2. **Formulate hypotheses:** - Null hypothesis $H_0$: $p = 0.7$ (the proportion is 70%) - Alternative hypothesis $H_a$: $p \neq 0.7$ (the proportion is different from 70%) 3. **Given data:** - Sample size $n = 100$ - Number owning laptops $x = 62$ - Sample proportion $\hat{p} = \frac{62}{100} = 0.62$ - Significance level $\alpha = 0.01$ 4. **Test statistic formula:** For testing a population proportion, $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where $p_0 = 0.7$ is the claimed proportion. 5. **Calculate the test statistic:** $$ z = \frac{0.62 - 0.7}{\sqrt{\frac{0.7 \times 0.3}{100}}} = \frac{-0.08}{\sqrt{0.0021}} = \frac{-0.08}{0.0458} \approx -1.75 $$ 6. **Determine critical value:** For a two-tailed test at $\alpha=0.01$, the critical z-values are approximately $\pm 2.576$. 7. **Decision:** Since $-1.75$ is between $-2.576$ and $2.576$, we fail to reject the null hypothesis. 8. **Interpretation:** There is not enough evidence at the 1% significance level to conclude that the proportion of students owning laptops differs from 70%.