The Confidence Interval of the Mean - Analysis Essay

Among the most important aspect of statistics is to help in the analysis of a given set of data. Using information obtained from a given sample population, a researcher is able to come up with inferences about a particular population. For this reason, statistics finds wide application in various disciplines. In light of this realization, this exercise aims at measuring the views that customers have on the ranges of bills for a party of two people for lunch and dinner at a particular restaurant. The study will employ some methods of data collection where information from the sample population will basically help in the understanding of the views that everyone in the population has on the particular subject under study. To achieve this, the method of confidence intervals was used to facilitate the analysis of the information obtained.

Analysis

Based on the information obtained about the prices of lunch and dinner at the restaurant, it was prudent to identify the confidence interval of the mean. This was based on the use of an alpha value of 0.05 hence an indication of 95% confidence level.

Table 1 below showing the prices of lunch and dinner at the restaurant

Lunch Dinner

$23.36 $44.00

$12.50 $51.90

$15.87 $49.70

$17.56 $40.00

$14.56 $55.50

$19.22 $33.00

$22.00 $43.40

$10.15 $41.30

$18.50 $45.20

$15.20 $40.70

$20.90 $41.10

$12.10 $49.10

$18.80 $30.90

$13.10 $45.20

$19.20 $55.30

$18.60 $52.10

$15.50 $55.10

$13.30 $38.80

$11.90 $43.10

$14.00 $39.20

$15.20 $58.60

$17.90 $49.80

$16.60 $43.20

$19.20 $47.90

$18.80 $46.60

Lunch

Based on the information obtained in the table, the following could be calculated for the lunch:

Sum = $414.02

Average or Mean = $414.02/25 = $16.5608

Sum of differences = $274.0722

Standard deviation= ($274.0722) / (25-1) = $3.3793

Dinner

Based on the information obtained in the table, the following could be calculated for the dinner:

Sum = $1140.7

Mean = $1140.7/25 = $45.628

Sum of differences = $1162.1504

s= ( $1162.1504) / (25-1) = $6.9587

Lunch

= approximate standard error of the mean formula

It equals to $3.3793/25 = $0.676

a/2 = 1-0.95 = 0.0025

t =0.025 = 2.064

Dinner

= approximate standard error of the mean formula

It equals to $6.9587/25 = $1.392

Lunch

UCL(mean) = $16.5608 + (2.064)($0.676) = $17.956

LCL(mean) = $16.5608 - (2.064)($0.676) = $15.166

Dinner

Upper class limit (Mean) = $45.628 + (2.064)*($1.392) = $48.501

Lower class limit (mean) = $45.628 (2.064)*($1.392) = $42.755

Conclusion

Using the information obtained from the calculations made above; there are some important inferences that can be made about the population under study. The confidence interval results show that the lunch prices range from approximately $15.166 to $17.956 for a dinner party arranged for two people. The results imply that 95% of all the customers representing the population under study should pick lunch for two ranging from $15.166 to $17.956. The results also show that the bill for dinner for two at the restaurant should range from $42.755 to $48.501. Therefore, it is prudent that the restaurant should set their bills within the range since the views of the majority of its customers believe the range is fair. The results have been tested at 95% confidence interval and hence represent the opinion and views of the majority of the customers view on the issue of price ranges for lunch and dinner for two at the restaurant.