Cl All P-P 335 1196' pq n (Round to two decimal places as needed.) Evaluate the formula z = Z= when n=1196, p=0.30, and q=1 - p. www
Cl All P-P 335 1196' pq n (Round to two decimal places as needed.) Evaluate the formula z = Z= when n=1196, p=0.30, and q=1 - p. www
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Evaluating the Z-Scores in Hypothesis Testing**
In this section, we will explore how to evaluate the formula for the z-score, which is a measure of how many standard deviations an element is from the mean. The formula we will use is:
\[ z = \frac{\hat{p} - p}{\sqrt{\frac{pq}{n}}} \]
where:
- \(\hat{p}\) is the sample proportion.
- \(p\) is the population proportion.
- \(q\) is \(1 - p\).
- \(n\) is the sample size.
Let's evaluate this formula with the specific values provided:
- \(\hat{p} = \frac{335}{1196}\)
- \(n = 1196\)
- \(p = 0.30\)
- \(q = 1 - p\)
First, we compute \(\hat{p}\):
\[ \hat{p} = \frac{335}{1196} \approx 0.28 \]
Next, we plug in these values into the z-score formula:
\[ z = \frac{0.28 - 0.30}{\sqrt{\frac{0.30 \times 0.70}{1196}}} \]
This simplifies to:
\[ z = \frac{-0.02}{\sqrt{\frac{0.21}{1196}}} \]
Calculate the denominator:
\[ \sqrt{\frac{0.21}{1196}} \approx \sqrt{0.0001755} \approx 0.01324 \]
Now, simplify the z-score:
\[ z = \frac{-0.02}{0.01324} \approx -1.51 \]
So, using the z-score formula:
\[ z \approx -1.51 \]
**Conclusion:**
To find the z-score, substitute the given values, perform the arithmetic operations, and round the final z-score to two decimal places if necessary. In this example, the z-score is approximately \(-1.51\), rounded to two decimal places.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc18b90ea-cb53-41e6-9532-f9af83d8e951%2F1e348e92-fd88-4f7a-a550-e8903ace1e37%2Fph0s6ig_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Evaluating the Z-Scores in Hypothesis Testing**
In this section, we will explore how to evaluate the formula for the z-score, which is a measure of how many standard deviations an element is from the mean. The formula we will use is:
\[ z = \frac{\hat{p} - p}{\sqrt{\frac{pq}{n}}} \]
where:
- \(\hat{p}\) is the sample proportion.
- \(p\) is the population proportion.
- \(q\) is \(1 - p\).
- \(n\) is the sample size.
Let's evaluate this formula with the specific values provided:
- \(\hat{p} = \frac{335}{1196}\)
- \(n = 1196\)
- \(p = 0.30\)
- \(q = 1 - p\)
First, we compute \(\hat{p}\):
\[ \hat{p} = \frac{335}{1196} \approx 0.28 \]
Next, we plug in these values into the z-score formula:
\[ z = \frac{0.28 - 0.30}{\sqrt{\frac{0.30 \times 0.70}{1196}}} \]
This simplifies to:
\[ z = \frac{-0.02}{\sqrt{\frac{0.21}{1196}}} \]
Calculate the denominator:
\[ \sqrt{\frac{0.21}{1196}} \approx \sqrt{0.0001755} \approx 0.01324 \]
Now, simplify the z-score:
\[ z = \frac{-0.02}{0.01324} \approx -1.51 \]
So, using the z-score formula:
\[ z \approx -1.51 \]
**Conclusion:**
To find the z-score, substitute the given values, perform the arithmetic operations, and round the final z-score to two decimal places if necessary. In this example, the z-score is approximately \(-1.51\), rounded to two decimal places.
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