Question

*NEED HELP PLEASE! ALSO Please show you got the Standard deviation to get the margin of error!**

1. To summarize the data, compute the proportion of all clients that a. Answered "Poor"to room quality b. Answered "Poor"to food quality C. Answered "Poor" to service quality 2. Develop the 92% confidence interval for the proportion of all recent clients that were "dissatisfied." What is the margin of error? What are the lower (or left) and upper (or right) endpoints of the confidence interval? Interpret your results 3. Develop the 92% confidence interval for the proportion of all recent clients who answered "Poor" to room quality. What is the margin of error? What are the lower (or left) and upper (or right) endpoints of the confidence interval? Interpret your results 4. Develop the 92% confidence interval for the proportion of all recent clients who answered "Poor" to food quality. What is the margin of error? What are the lower (or left) and upper (or right) endpoints of the confidence interval? Interpret your results 5. Develop the 92% confidence interval for the proportion of all recent clients who answered "Poor" to service quality. What is the margin of error? What are the lower (or left) and upper (or right) endpoints of the confidence interval? Interpret your results 6. Conduct a hypothesis test (using either the p-Value Approach or the Critical Value Approach) to determine if the proportion of all recent clients is more dissatisfied than the traditional level of dissatisfaction. Use α = 0.08. Do not forget to include the correctly worded hypotheses and show all the steps required to conduct the hypothesis test. 7.What advice would you give to A1 Hotels based upon your analysis of the data? What is the magnitude of the issue? How can this study be improved? GGGGGGGGPGGPGGPPGGP Qradir PGGPGGGPGGGPGGGPGGG 0a GGGGGPGGPGGPGGGPGGP 1234567890123456789 GGGPGGPPGPGGPGGGGGGGP PGPPGGGGGPPGPPPGPGPGG 012345678901234567890 222222222233333333334 GGPPGGPGPPGGPGGGPGPPP PGGGGGPGPPGPPGGGGGPGP GGGPPPGGGPGPPGGGGGGPG 123456 7 89012345678901 444445555555555 66 62 G 63 G 64G 65 P 66 G 67 G 68 G 69 G 70 G 71 G 72 G 73 G 74 G 75G 76 P 77 G 78 G 79 G 80 G 81G 82 G GGGPGPGPGGGGPGPGGGPPG GGGPGPGGGGPGPGGPPGPGG pGGGGPGPGGGGGGPGPGGGG 34567 8901234567890123 8 8 8 8 888 999 99 99999 0 0 0 0 GGGG G P G G G G G G G G P G G G G G 012345678901234 GGGGPGPGGGGGGGGGGGGPG GGGGGGGPGGGGGPPGPGGPG GGGGGGGGGGGGGGPGPGGPG 56789 012345678901 23 22222333333333( 4 5 34444 146 G 147 G 148 G 149 G 150 G 151 G 152 G 153 G 154 G 155 G 156 G 157 P 158 G 159 G 160 G 161 G 162 P 163 G 164 G 165 P 166 G PGGPGPGGGGGGGPGGGGPGG GGGGGPGGPGGPGGGGGGPGG GPGGGPGGGGGPGGGGGGGGG 7, 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 188 G 189G 190 G 191 G 192 G 193 G 194 G 195 G 196 G 197 G 198 G 199G 2 200 G

Answer 1

Since there are lot of data points

I am going to tell you the formulas

X =number of success

n = total sample size

For example if we want to get confidence interval for those who answered poor to room quality

X = number of person who answered poor to room quality

now

p^ = X/n

here n = 200

say

X = 120 {note that I am just taking a guess, count yourself using Excel formulas}

then

p^ = X/n = 0.6

now

margin of error = z *sqrt(pq/n)

here z depends on confidence level

for 92% confidence level

z = normsinv(0.92 + (1- 0.92)/2) = 1.7507

hence

margin of error

= 1.7507 * sqrt(0.6 * (1-0.6)/200)

= 0.0606

confidence interval is

(p^ - margin of error , p^ + margin of error)

=(0.6- 0.0606 , 0.6 + 0.0606 )

= ( 0.5394 , 0.6606 )

Please rate

Please note again that I took X as 120 randomly if you calculate yourself and repeat the process

Thanks