What is the critical value for the following scenario?

Hilda is a quality assurance manager. The product (bag of chips) is advertised as 500 gr of chips in each bag. She has taken a sample of 49 bags of chips and has calculated the average as 512.5 gr. The sample's standard deviation is 3.45 gr. She needs to construct the 95% confidence interval.

**enter the positive value only**


To find the critical value for constructing the 95% confidence interval, the sample size is 49, which is greater than 30. Hence, we can use the Z-distribution as an approximation.

For a 95% confidence interval:
- The confidence level is 95%, so the alpha level (significance level) is \(1 - 0.95 = 0.05\).
- Since it's a two-tailed test, divide the alpha by 2: \(\frac{0.05}{2} = 0.025\).

Now, we need to find the Z-score that corresponds to the upper 2.5% of the normal distribution (as it's a two-tailed test with a 95% confidence level). This Z-score, known as the critical value, is:

\[
Z_{0.025} = 1.96
\]

So, the critical value for this scenario is:

\[
\boxed{1.96}
\]

Would you like a detailed explanation of how this was calculated or have any further questions?

Here are some follow-up questions related to confidence intervals and hypothesis testing:

1. How do you use the critical value to construct the confidence interval for the mean?
2. Why is the Z-distribution appropriate for this sample size?
3. How would the confidence interval change if a different confidence level was used, such as 99%?
4. What does the confidence interval tell us about the population mean?
5. How would the result differ if the sample standard deviation was significantly larger?

**Tip:** For large sample sizes (typically \( n > 30 \)), the sample mean approximates a normal distribution regardless of the population's distribution, thanks to the Central Limit Theorem.

What is the critical value for the following scenario? Hilda is a quality assurance manager. The product (bag of chips) is advertised as 500 gr of chips in each bag. She has taken a sample of 49 bags of chips and has calculated the average as 512.5 gr. The sample's standard deviation is 3.45 gr. She needs to construct the 95% confidence interval. **enter the positive value only**

Learn how to find the critical value for a 95% confidence interval in a quality assurance scenario with 49 sample bags of chips. Understand the Z-distribution, Central Limit Theorem, and confidence interval formula. The sample's average weight is 512.5 gr with a standard deviation of 3.45 gr.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation