b. What is the test statistic?c. What is the P-value?d. Identify the null hypothesis.e. What is the final conclusion?A.There is sufficient evidence to support the claim that 94% of adults own a cell phone.B.There is sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone.C.There is not sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone.D.There is not sufficient evidence to support the claim that 94% of adults own a cell phone. A poll of 2,085 randomly selected adults showed that 93% of them own cell phones. The technology display below results from a test of the claim that 95% of adultsown 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 p = 0.95 vs p#0.95Sample XNSample p95% CIZ-ValueP-Value119422,0850.931415(0.917157,0.945673)- 3.890.000a. Is the test two-tailed, left-tailed, or right-tailed?O Right tailed testLeft-tailed testTwo-tailed test Choose the correct answer below.O A. Reject the null hypothesis because the P-value is less than or equal to 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 greater than the significance level, a.D. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.

Name: b. What is the test statistic?c. What is the P-value?d. Identify the
Uploaded: 2024-03-20T07:07:12.000Z
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Description: b. What is the test statistic?c. What is the P-value?d. Identify the null hypothesis.e. What is the final conclusion?A.There is sufficient evidence to support the claim that 94% of adults own a cell phone.B.There is sufficient evidence to warrant

Since we only answer up to 3 sub-parts, I'll the first 3. Please resubmit the question and specify the other subparts. The given data is, α=0.05 p^=93% =0.93, n=2085 a) The given claim is the proportion is 95% i.e. 0.95. And the test is given for p is, p = 0.95 VS p ≠ 0.95 So, this test is two-tailed. b) The test statistic is, Z= p^-pp*(1-p)n=0.9319-0.950.95*(1-0.95)2085=~-3.89 where, p^=XN=0.9319c) The p-values in given in the question which is, p-value =0.000