A wedding website states that the average cost of a wedding is $26,706$⁢26,706. One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 3535 recently married couples and finds that the average cost of weddings in the sample was $24,365$⁢24,365. Assuming that the population standard deviation is $6575$⁢6575, is there sufficient evidence to support the bride’s hope at the 0.050.05 level of significance?

Step 3 of 3 :  

Draw a conclusion and interpret the decision.

### Step 3: Draw a Conclusion and Interpret the Decision

Now that we have the test statistic value from the previous step, which was approximately $-2.11$, we need to compare this value to the critical value at the $0.05$ level of significance to make a decision.

### Critical Value for a One-Tailed Test

Since the test is one-tailed (because the bride hopes that the average cost is **less than** the reported value), we need to find the critical z-value for a significance level of $0.05$.

For a left-tailed test at the $0.05$ significance level, the critical z-value is approximately:

\[
z_{0.05} = -1.645
\]

### Decision Rule

- **Reject the Null Hypothesis (H₀)** if the test statistic $z$ is less than the critical value ($-1.645$).
- **Fail to Reject the Null Hypothesis** if the test statistic $z$ is greater than or equal to the critical value ($-1.645$).

### Comparison

- Calculated test statistic: $-2.11$
- Critical value: $-1.645$

Since $-2.11 < -1.645$, we **reject the null hypothesis (H₀)**.

### Conclusion

There is sufficient evidence at the $0.05$ level of significance to support the bride's hope that the average cost of a wedding is less than $26,706.

Would you like any further details or have any questions?

Here are five related questions you might find helpful:
1. What does rejecting the null hypothesis imply in this context?
2. How would the conclusion change if the significance level was set to 0.01?
3. What is the p-value associated with the test statistic, and how does it compare to the significance level?
4. How does the choice of a one-tailed test affect the conclusion?
5. What are the possible consequences of a Type I error in this context?

**Tip:** In hypothesis testing, rejecting the null hypothesis implies that the sample provides strong enough evidence to support the alternative hypothesis.

Explore a hypothesis testing example to determine if the average wedding cost is below $26,706. Analyze a sample data set with a population standard deviation of $6575 to make a decision at a 0.05 significance level.

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