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…

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