Answer

Introduction

A confidence interval is a type of interval estimate, computed from the statistics of the observed data that might contain the true value of an unknown population parameter.

It can be used to make decisions and predictions about the issue under study.

The data provided in this essay is about A1 hotels that operates luxury hotels throughout the worlds. A1 hotels would like to study customer satisfactions so they conducted a survey to analyze the customer satisfaction of people visiting the hotels.

The report below shows the responses that was obtained from the 200 surveys that were taken at random.

The proportion of all clients that

Answered “Poor” to room quality. = (52/200) =26% a total of 52 clients out of 200 gave their response to room quality as poor. This is a 26% proportion of the total responses from 200 surveys.

Answered “Poor” to food quality.= (55/200) = 27.5%. The total number of responses were 200 but out of the 200, 55 answered poor to food quality. This is a percentage proportion of 27.7%.

Answered “Poor” to service quality.= (60/200) = 30%. 60 people out of 200 responded poor to service quality. This makes a 30% proportion of the total responses.

 92% confidence interval for the proportion of all recent clients that were “dissatisfied.”

The total number of Dissatisfied clients = 91 out of 200. This are those users that answered poor to either food quality, service quality or room quality.

This entails a percentage of 45.5% of the total surveys taken. (91/200)*100 = 45.5%

Z index of 92% = 1.75

Where p is the percentage.

0.455(1-0.455)/200 = 0.455*0.545/200 = 0.247975/200 = 0.001239875

√0.001239875 = 0.035211858797

The marginal error is 1.75 * 0.035211858797 = 0.06162075289475 =6.1%

Marginal error = 6.1 %

Lower limit = 45.5% – 6.1% =39.4%

Upper limit = 45.5% + 6.1% = 51.7 %

The range is 51.7 – 39.4 =12.2%

This means that the approximate percentage of customers that are dissatisfied ranges from 39.4% to %51.7% of the total population of the customers visiting the hotel.

 92% confidence interval for the proportion of all recent clients who answered poor to room quality.

Total number of customers that answered poor to room quality is 52 out of 200.  The percentage proportion is = (52/200)*100 = 26%

Z index of 92% = 1.75

0.26(0.74)/200 = 0.000962

 √0.000962 = 0.0310161248385416 = 3.1%

Marginal error = 3.1%

Lower limit = 26 – 3.1 = 22.9%

Upper limit = 26 + 3.1 = 29.1%

This means in any random 200 surveys responses the number of people that will answer poor to room quality will range from 22.9% and 29.1%.

 92% confidence interval for the proportion of all recent clients who answered poor to food quality.

Total number of customers that answered poor to food quality is 55 out of 200.  The percentage proportion is = (55/200)*100 = 27.5%

Z index of 92% = 1.75

0.275(0.725)/200 = 0.000996875

 √0.000962 = 0.0315733273507877= 3.2%

Marginal error = 3.2%

Lower limit = 27.5 – 3.1 = 24.4%

Upper limit = 27.5 + 3.1 = 30.6%

This means in any random 200 surveys responses the number of people that will answer poor to food quality will range from 24.4% and 30.6%.

 92% confidence interval for the proportion of all recent clients who answered poor to service quality.

Total number of customers that answered poor to service quality is 60 out of 200.  The percentage proportion is = (60/200)*100 = 30%

Z index of 92% = 1.75

0.30(0.70)/200 = 0.00105

 √0.000962 = 0.0324037034920393= 3.2%

Marginal error = 3.2%

Lower limit = 30 – 3.2 = 26.8%

Upper limit = 30 + 3.2 = 33.2%

This means in any random 200 surveys responses the number of people that will answer poor to room quality will range from 26.8% and 33.2%.

Hypothesis using the p value approach: p value hypothesis

H_o: p = p_o
H_a: p ≠ p_o 

Let the percentage of dissatisfied be p. Null hypothesis: p = 40% and Alternative hypothesis: p > 40%

Z = (0.455 – 0.4) / (√ (0.4(1-0.4)/200)

0.055 / 0.0346410161513775 = 1.587713240271473

Z = 1.59

Using the standard normal table we get the value to be 0.0559

From the standard normal table the value is 0.0559 and it is less than 0.08 hence we drop the null hypothesis and decide in favor of the alternative hypothesis.

We conclude that the proportion of dissatisfied people visiting the hotel is higher than 40%.

A1 hotel will be advised to improve the services offered to its customers in order to get a favorable response.

The magnitude of the issue is high since the number of dissatisfied people is increasing from 40%. Action need to be taken to improve the food quality, room quality and service quality. Service quality needs to be improved because it has the highest proportion of people that answered poor to its quality. Also food quality need to be improved and a smaller improvement applied to room quality,

The study can be improved by providing a scale where the customers can give their level of satisfaction and the reason for their response in order for the improvement to be done. A response scale of between 1 and 10 would work perfectly to provide customers with a wide range of expression on how they found the services of the hotel.

Conclusion

From the above information, it can be concluded that the number of dissatisfied customers is increasing from 40% to a higher value and immediate action needs to be taken to reduce this percentage dissatisfaction.

Also it can be concluded that service quality has the highest number of customers who answered poor and something needs to be done to improve it.

Finally, a survey of a scale of 1 to10 can be used to give more room for customers to express the quality of services offered to them.