Let's answer the questions based on the given data and information provided.

1. Level of significance

The level of significance ($\alpha$) is not explicitly provided in the data, but it is typically assumed to be 0.05 for a 95% confidence level. Thus:
$\alpha = 0.05$

2. Null hypothesis ($H_0$)

The null hypothesis states that the proportion of customers who make a purchase is equal to 50%. Mathematically:
$H_0: p = 0.5$

3. Alternative hypothesis ($H_a$)

The alternative hypothesis states that the proportion of customers who make a purchase is different from 50%. This is a two-tailed test. Mathematically:
$H_a: p \neq 0.5$

4. Should $H_0$ be rejected or does $H_0$ fail to be rejected?

To determine this, we compare the p-value to the level of significance ($\alpha$):

• Given p-value = 0.072
• Given $\alpha = 0.05$

Since the p-value (0.072) is greater than the level of significance ($\alpha = 0.05$), $H_0$ fails to be rejected.

5. Conclusion

At the 5% level of significance, there is insufficient evidence to conclude that the proportion of customers who make a purchase is different from 50%. Therefore, the observed difference is not statistically significant.

6. Population parameter

The population parameter of interest is the true proportion of all customers who visit the site and make a purchase ($p$).

7. Condition for using a z-distribution

The condition for using the z-distribution is based on the sample size being sufficiently large to approximate a normal distribution:

• $n \cdot p_0 \geq 10$
• $n \cdot (1 - p_0) \geq 10$

Here:

• $n = 100$
• $p_0 = 0.5$

Check the conditions:
$n \cdot p_0 = 100 \cdot 0.5 = 50 \quad \text{and} \quad n \cdot (1 - p_0) = 100 \cdot 0.5 = 50$
Both values are greater than 10, so the z-distribution condition is met.

Summary

1. Level of significance: $\alpha = 0.05$
2. Null hypothesis $H_0$: $p = 0.5$
3. Alternative hypothesis $H_a$: $p \neq 0.5$
4. Decision: $H_0$ fails to be rejected.
5. Conclusion: Insufficient evidence to conclude the proportion is different from 50%.
6. Population parameter: True proportion of customers making a purchase ($p$).
7. Condition met: $n \cdot p_0 \geq 10$ and $n \cdot (1 - p_0) \geq 10$.

Additional Questions to Explore:

1. What does the test statistic $z = 1.8$ indicate in terms of standard deviations?
2. How does the critical value ($1.95$) relate to the decision rule?
3. What would happen if the confidence level was 99% instead of 95%?
4. How could increasing the sample size influence the p-value?
5. What type of error might occur in this test (Type I or Type II)?

Tip:

In hypothesis testing, always compare the p-value to the level of significance to determine whether to reject $H_0$. A smaller p-value (< $\alpha$) provides stronger evidence against $H_0$.

Let me know if you'd like further details or clarification!