統計學report代寫:餐廳潛在客戶群

發布時間:2019-10-19 22:51
統計學report代寫:餐廳潛在客戶群

本報告旨在提供一個報告,介紹是否適合開一家精致精致的高檔餐廳,提供精致的菜肴、飲料和其他小吃,并以優雅的餐廳為背景。從這一點出發,介紹了餐廳的可能的顧客率,潛在客戶群,潛在客戶偏好的食物,一種娛樂方式,在一起withadvertising渠道等等。這些數據是從400份調查問卷提供的信息。該報告將重點討論餐廳正常運作和服務目標的10個關鍵問題。
2。對于主菜潛在的支付
潛在的支付每主菜可以提供客戶的需求、意愿的總的看法,和相對成本的餐廳可以提供食品級。
根據研究公司的預測模型,為點菜入口éE是較好的平均支付18美元。使用一個樣本測試來比較數據從已知值到已知值的平均值是很有用的。當數據只涉及一個樣本且種群方差明顯未知時,該方法也適用,且因變量是正態分布的。
在這個數據集中,根據下面的柱狀圖,平均付款變量是正偏的。然而,這些數據基本上是平均分布在18左右,而且這些點幾乎與基準線周圍的一條線相連,因此可以把變量視為正態分布,并且可以對其進行一樣本t-檢驗。
 
在這里,零假設平均熵éE支付是18美元,另一種假設是,付款不等于18美元。利用SPSS、單樣本t檢驗顯示如下。
 
一個樣本統計
N的平均標準偏差的標準誤差
你預期平均晚餐主菜項目單獨定價會怎樣?340 18.8353美元9.82784美元0.53299
 
 
一個樣品測試
測試值= 18
數字信號接口。(雙尾)的平均差異的95%置信區間的差異
上下
你預期平均晚餐主菜項目單獨定價會怎樣?1.567 339. 118美元0.83529 - 0.2131美元1.8837美元
 
平均值為18.8353,接近18,但標準偏差會影響平均值的穩定性,這是不夠的。幸運的是,這一個樣本的平均差異值為0.83529,P值為0.118,1.567and T值,大于0.05的臨界值,表明從18差異不顯著。同時,差異的95%置信區間,其范圍從-0.2131 1.8837,包括零。零假設不能被拒絕,樣本均值等于給定值。

統計學report代寫:餐廳潛在客戶群

This report is aimed to provide a report about whether it is suitable to open a fine and upscale restaurant which serves delicate dishes, beverages and other snacks with the setting of the restaurant being elegant. From this point, the report deals with the possible patron rate of the restaurant, the potential customer group, potential customers’ preference of the kind of the foods, entertainment style, together withadvertising channels and so on.The data is supplied information from 400 questionnaires. The report will be focused on 10 key problems concerning the normal operation and the serving target of the restaurant.
2. Potential Payment for the Entrees
Potential payment per entrees could offer a general view of the customers’ demand, willingness, and level of food with relative cost the restaurant could provide.
According to the forecasting model of the research firm, the preferable average payment for an a la carte entrée is $18. It is useful to use one-sample test to compare the mean of the data from answers to a known value. The method is suitable when the data only concerns one sample and the population variance is obviously unknown, as well as the dependent variable is normally distributed.
In this dataset, the variable of average payment is positively skewed according to the histogram below. However, the data is almost normally distributed around the average of 18 and the dots are nearly linked as a line around the benchmark line, therefore the variable could be regarded as normally distributed and the one-sample t-test on it could proceed. 
 
Here, the null hypothesis is that the average entrée payment is $18 and the alternative hypothesis is that the payment is not equal to $18. Using SPSS, the resultof the one-sample t-test is displayed as follows.
 
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
What would you expect an average evening meal entree item alone to be priced? 340 $18.8353 $9.82784 $0.53299
 
 
One-Sample Test
Test Value = 18
t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference
Lower Upper
What would you expect an average evening meal entree item alone to be priced? 1.567 339 .118 $0.83529 -$0.2131 $1.8837
 
The mean value is 18.8353, which is close to 18, but this is not enough as the standard deviation will influence the stability of the mean value. Fortunately, the mean difference value of the one-sample test is 0.83529, and the t value of 1.567and p value of 0.118, which is greater than the critical value of 0.05, indicates that the difference from 18 is not significant. Also, the 95% Confidence Interval of the Difference, which ranges from -0.2131 to 1.8837, includes zero. The null hypothesis could not be rejected and that the sample mean is equal to the given value.
如果您有論文代寫需求,可以通過下面的方式聯系我們

提交代寫需求

如果您有論文代寫需求,可以通過下面的方式聯系我們。

在線客服

售前咨詢
售后咨詢
微信號
Essay_Cheery
微信
英国代写_数学代写_c++/c代写_留学生代写怎么查出来?