Question

A poll of 2,024 randomly selected adults showed that 94​% of them own cell phones. The technology display below results from a test of the claim that 90​% of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.01

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

Test of pequals=0.9 vs p≠0.9
Sample	X	N	Sample p	​95% CI	​Z-Value	​P-Value
1	1904	2,024	0.940711	​(0.927190,0.954233)	6.11	0.000

a. Is the test​ two-tailed, left-tailed, or​ right-tailed?

a)Two-tailed test

b)Right tailed test

c)​Left-tailed test

b. What is the test​ statistic?

​(Round to two decimal places as​ needed.)

c. What is the​ P-value?

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

a)Ho:= p=0.9

b)Ho:p<0.9

c) Ho: p≠0.9

d)Ho: p>0.9

Answer 1

Given : Sample size=n=2024

The estimate of the sample proportion =0.94

Significance level=0.01

Hypothesized value=0.90

We have , Hypothesis: VS

a) The test is two-tailed test.

b)The test statistic is ,

c) The p-value is ,

p-value=

The Excel function is , =2*(1-NORMDIST(6,0,1,TRUE))

d) The null hypothesis is ,

Decision : Here , p-value=0.0000 <

Therefore , reject Ho.

Conclusion : Hence , there is not sufficient evidence to support the claim that the 90% of adults own cell phones.