A poll of 2 comma 035 randomly selected adults showed that 90% of them own cell phones. The technology display below results from a test of the claim that 91% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level.a. What is the test statistic?The test statistic is(Round to two decimal places as needed.) b. What is the P-value?The P-value is(Round to three decimal places as needed.) ### Statistical Test Summary**Hypothesis Test:**- **Null Hypothesis (H₀):** p = 0.91- **Alternative Hypothesis (H₁):** p ≠ 0.91**Sample Data:**- **Sample Number:** 1- **X (Number of Successes):** 1825- **N (Sample Size):** 2035**Sample Proportion:**- **Sample p:** 0.896806**Confidence Interval:**- **95% CI:** (0.879435, 0.914176)**Statistical Test Results:**- **Z-Value:** -2.08- **P-Value:** 0.038### Interpretation:- The test is conducted to determine if the proportion \( p \) is significantly different from 0.91.- The calculated z-value is -2.08, indicating how many standard deviations the sample proportion is from the hypothesized population proportion.- A p-value of 0.038 suggests that there is a statistically significant difference at the common alpha level of 0.05, meaning we reject the null hypothesis and conclude that the true proportion \( p \) is likely not 0.91.- The 95% confidence interval provides a range where the true proportion is expected to fall, based on the sample data. In this case, the interval spans from 0.879435 to 0.914176.

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Answered: A poll of 2 comma 035 randomly selected…

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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A poll of 2 comma 035 randomly selected adults showed that 90% of them own cell phones. The technology display below results from a test of the claim that 91% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level.

a. What is the test statistic?

The test statistic is

(Round to two decimal places as needed.)

b. What is the P-value?

The P-value is

(Round to three decimal places as needed.)

### Statistical Test Summary

**Hypothesis Test:**
- **Null Hypothesis (H₀):** p = 0.91
- **Alternative Hypothesis (H₁):** p ≠ 0.91

**Sample Data:**
- **Sample Number:** 1
- **X (Number of Successes):** 1825
- **N (Sample Size):** 2035

**Sample Proportion:**
- **Sample p:** 0.896806

**Confidence Interval:**
- **95% CI:** (0.879435, 0.914176)

**Statistical Test Results:**
- **Z-Value:** -2.08
- **P-Value:** 0.038

### Interpretation:

- The test is conducted to determine if the proportion \( p \) is significantly different from 0.91.
- The calculated z-value is -2.08, indicating how many standard deviations the sample proportion is from the hypothesized population proportion.
- A p-value of 0.038 suggests that there is a statistically significant difference at the common alpha level of 0.05, meaning we reject the null hypothesis and conclude that the true proportion \( p \) is likely not 0.91.
- The 95% confidence interval provides a range where the true proportion is expected to fall, based on the sample data. In this case, the interval spans from 0.879435 to 0.914176.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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