A poll of2,088randomly selected adults showed that96%of them own cell phones. The technology display below results from a test of the claim that95%of adults own cell phones. Use the normal distribution as an approximation to the binomialdistribution, and assume a0.01significance level to complete parts (a) through (e). Test ofp=0.95vsp≠0.95SampleXNSample p95% CIZ-ValueP-Value119972,0880.956418(0.944909,0.967926)1.350.178 a. Is the test two-tailed, left-tailed, or right-tailed?O Right tailed testTwo-tailed testO Left-tailed testb. What is the test statistic?The test statistic is.(Round to two decimal places as needed.)c. What is the P-value?The P-value is(Round to three decimal places as needed.) Identify the null hypothesis.О А. Но: р> 0.95О В. Но: р<0.95О с. Но: р30.95O D. Ho: p#0.95Choose the correct answer below.O A. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.O B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a.OC. Reject the null hypothesis because the P-value is less than or equal to the significance level, a.O D. Reject the null hypothesis because the P-value is greater than the significance level, a.e. What is the final conclusion?O A. There is sufficient evidence to support the claim that 95% of adults own a cell phone.

a) The hypotheses are given below: Null hypothesis: H0 :p =0.95. Alternative hypothesis: Ha :p ≠0.95…

Answered: A poll of 2,088 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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Question

A poll of

2,088

randomly selected adults showed that

96%

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 binomialdistribution, and assume a

0.01

significance level to complete parts (a) through (e).

Sample	X	N	Sample p	95% CI	Z-Value	P-Value
Test of p=0.95 vs p≠0.95
1	1997	2,088	0.956418	(0.944909,0.967926)	1.35	0.178

a. Is the test two-tailed, left-tailed, or right-tailed?
O Right tailed test
Two-tailed test
O Left-tailed test
b. What is the test statistic?
The test statistic is.
(Round to two decimal places as needed.)
c. What is the P-value?
The P-value is
(Round to three decimal places as needed.)

Identify the null hypothesis.
О А. Но: р> 0.95
О В. Но: р<0.95
О с. Но: р30.95
O D. Ho: p#0.95
Choose the correct answer below.
O A. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.
O B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a.
OC. Reject the null hypothesis because the P-value is less than or equal to the significance level, a.
O D. Reject the null hypothesis because the P-value is greater than the significance level, a.
e. What is the final conclusion?
O A. There is sufficient evidence to support the claim that 95% of adults own a cell phone.

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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