A poll of 2,018 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 95% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e).Is the test two-tailed, left-tailed, or right-tailed?Left-tailed testRight tailed testTwo-tailed test What is the test statistic?The test statistic is(Round to two decimal places as needed.) What is the P-value?(Round to three decimal places as needed.) What is the null hypothesis and what do you conclude about it?Identify the null hypothesisHo: p> 0.95Ho: p= 0.95Ho: p< 0.95Ho: p≠ 0.95Choose the correct answer below.Fail the null hypothesis because the P-value is greater than the significance level, αReject the null hypothesis because the P-value is less than or equal to the significance level, αReject the null hypothesis because the P-value is greater than the significance level, αFail to reject the null hypothesis because the P-value is less than or equal to the significance What is the final conclusion?There is not sufficient evidence to support the claim that 95% of adults own a cell phone.There is sufficient evidence to warrant rejection of the claim that 95% of adults own a cell phone.There is not sufficient evidence to warrant rejection of the claim that 95% of adults own a cell phone.There is sufficient evidence to support the claim that 95% of adults own a cell phone. Test of p = 0.95 vs p#0.95Sample XSample p95% CIZ-ValueP-Value118892,0180.936075(0.925403,0.946748)- 2.870.004

Answered: A poll of 2,018 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 2,018 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 95% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e).

Is the test two-tailed, left-tailed, or right-tailed?

Left-tailed test

Right tailed test

Two-tailed test

What is the test statistic?

The test statistic is

(Round to two decimal places as needed.)

What is the P-value?

(Round to three decimal places as needed.)

What is the null hypothesis and what do you conclude about it?

Identify the null hypothesis

Ho: p> 0.95
Ho: p= 0.95
Ho: p< 0.95
Ho: p≠ 0.95

Choose the correct answer below.

Fail the null hypothesis because the P-value is greater than the significance level, α
Reject the null hypothesis because the P-value is less than or equal to the significance level, α
Reject the null hypothesis because the P-value is greater than the significance level, α
Fail to reject the null hypothesis because the P-value is less than or equal to the significance

What is the final conclusion?
There is not sufficient evidence to support the claim that 95% of adults own a cell phone.
There is sufficient evidence to warrant rejection of the claim that 95% of adults own a cell phone.
There is not sufficient evidence to warrant rejection of the claim that 95% of adults own a cell phone.
There is sufficient evidence to support the claim that 95% of adults own a cell phone.

Test of p = 0.95 vs p#0.95
Sample X
Sample p
95% CI
Z-Value
P-Value
1
1889
2,018
0.936075
(0.925403,0.946748)
- 2.87
0.004

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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