1 TIMEREMAINING $$ 46: 43 $$ Julian is a manager at a clothing store for teens. He is analyzing the order for next seas̈on. Data for the previous 10 years suggests that teens are willing to spend an average of $$ \$ 75 $$ for a pair of designer jeans with a standard deviation of $$ \$ 5 $$. However, Julian thinks the average may have changed due to a recession. He finds that the last three seasons of data show that teens spent an average of $$ \$ 68 $$ on a pair of jeans. Therefore, he performed a hypothesis test to see if the recent average is the same. Julian used a significance level of $$ 5 \% $$ to perform the test. Which of the following statements is valid based on the results? Julian's data shows that the recent seasons' average jean price is not $$ \$ 75 $$. Julian's data shows that he should order fewer jeans than before. Julian's data shows that the recent seasons' average jean price is still $$ \$ 75 $$. Julian's data is inconclusive, so he should order the same number of jeans.

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To analyze Julian's hypothesis test, we need to follow these steps:

1. **Identify the null and alternative hypotheses:**
   - Null Hypothesis ($$H_0$$): The average price teens are willing to spend on jeans is $$ \$ 75 $$ (i.e., $$ \mu = 75 $$).
   - Alternative Hypothesis ($$H_a$$): The average price teens are willing to spend on jeans is not $$ \$ 75 $$ (i.e., $$ \mu \neq 75 $$).

2. **Determine the significance level:**
   - The significance level ($$\alpha$$) is $$ 0.05 $$ (5%).

3. **Calculate the test statistic:**
   - We will use the formula for the z-test since we know the population standard deviation:
   $$
   z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
   $$
   where:
   - $$\bar{x} = 68$$ (sample mean),
   - $$\mu = 75$$ (population mean under the null hypothesis),
   - $$\sigma = 5$$ (population standard deviation),
   - $$n$$ is the sample size. Since the sample size is not provided, we will assume it is large enough for the z-test to be valid.

4. **Determine the critical z-values:**
   - For a two-tailed test at the $$0.05$$ significance level, the critical z-values are approximately $$-1.96$$ and $$1.96$$.

5. **Make a decision:**
   - If the calculated z-value falls outside the range of $$-1.96$$ to $$1.96$$, we reject the null hypothesis.

Now, let's calculate the z-value using the provided data. Since we don't have the sample size, we will proceed with the calculation assuming a large sample size.

Let's calculate the z-value.

Given:
- $$\bar{x} = 68$$
- $$\mu = 75$$
- $$\sigma = 5$$

We will assume $$n$$ is sufficiently large for the calculation.

Now, let's calculate the z-value.

$$
z = \frac{68 - 75}{\frac{5}{\sqrt{n}}}
$$

Since we don't have $$n$$, we will express the z-value in terms of $$n$$:

$$
z = \frac{-7}{\frac{5}{\sqrt{n}}} = -\frac{7\sqrt{n}}{5}
$$

Now, we need to analyze the implications of this z-value.

### Conclusion:
- If $$n$$ is large enough such that $$z < -1.96$$, we reject $$H_0$$ and conclude that the average price is not $$ \$ 75 $$.
- If $$z$$ is between $$-1.96$$ and $$1.96$$, we fail to reject $$H_0$$ and conclude that there is not enough evidence to say the average price is different from $$ \$ 75 $$.

Based on the information provided, if Julian's data shows that the recent seasons' average jean price is not $$ \$ 75 $$, he should order fewer jeans than before.

Thus, the valid statement based on the results is:
**Julian's data shows that the recent seasons' average jean price is not $$ \$ 75 $$.**
Julian's data shows that the recent seasons' average jean price is not \$75.

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