统计学report代写:餐厅潜在客户群

统计学report代写:餐厅潜在客户群

本报告旨在提供一个报告，介绍是否适合开一家精致精致的高档餐厅，提供精致的菜肴、饮料和其他小吃，并以优雅的餐厅为背景。从这一点出发，介绍了餐厅的可能的顾客率，潜在客户群，潜在客户偏好的食物，一种娱乐方式，在一起withadvertising渠道等等。这些数据是从400份调查问卷提供的信息。该报告将重点讨论餐厅正常运作和服务目标的10个关键问题。
2。对于主菜潜在的支付
潜在的支付每主菜可以提供客户的需求、意愿的总的看法，和相对成本的餐厅可以提供食品级。
根据研究公司的预测模型，为点菜入口éE是较好的平均支付18美元。使用一个样本测试来比较数据从已知值到已知值的平均值是很有用的。当数据只涉及一个样本且种群方差明显未知时，该方法也适用，且因变量是正态分布的。
在这个数据集中，根据下面的柱状图，平均付款变量是正偏的。然而，这些数据基本上是平均分布在18左右，而且这些点几乎与基准线周围的一条线相连，因此可以把变量视为正态分布，并且可以对其进行一样本t-检验。

在这里，零假设平均熵éE支付是18美元，另一种假设是，付款不等于18美元。利用SPSS、单样本t检验显示如下。

一个样本统计
N的平均标准偏差的标准误差
你预期平均晚餐主菜项目单独定价会怎样？340 18.8353美元9.82784美元0.53299


一个样品测试
测试值= 18
数字信号接口。（双尾）的平均差异的95%置信区间的差异
上下
你预期平均晚餐主菜项目单独定价会怎样？1.567 339. 118美元0.83529 - 0.2131美元1.8837美元

平均值为18.8353，接近18，但标准偏差会影响平均值的稳定性，这是不够的。幸运的是，这一个样本的平均差异值为0.83529，P值为0.118，1.567and T值，大于0.05的临界值，表明从18差异不显著。同时，差异的95%置信区间，其范围从-0.2131 1.8837，包括零。零假设不能被拒绝，样本均值等于给定值。

统计学report代写:餐厅潜在客户群

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.