Related questions with answers

One gram of fat has nine calories. If you know the number of fat grams in a particular food, you can use the following formula to calculate the number of calories that come from fat in that food:

Calories from fat = Fat grams * 9

If you know the food’s total calories, you can use the following formula to calculate the percentage of calories from fat:

Percentage of calories from fat = Calories from fat /Total calories

Create an application that allows the user to enter:

The total number of calories for a food item

The number of fat grams in that food item

The application should calculate and display:

The number of calories from fat

The percentage of calories that come from fat

Also, the application’s form should have a CheckBox that the user can check if he or she wants to know whether the food is considered low fat. (If the calories from fat are less than 30% of the total calories of the food, the food is considered low fat.) Use the following test data to determine if the application is calculating properly:

 Calories and Fat                         Percentage Fat
200 calories, 8 fat grams       Percentage of calories from fat: 36%
150 calories, 2 fat grams       Percentage of calories from fat: 12% (a low-fat food)
500 calories, 30 fat grams      Percentage of calories from fat: 54%

Note: Make sure the number of calories and fat grams are not less than 0. Also, the number of calories from fat cannot be greater than the total number of calories. If that happens, display an error message indicating that either the calories or fat grams were incorrectly entered.

Question

Consider the data set 7, 9, 14, 15, 22. The standard error of the sample mean is: A. 2.53 B. 2.56 C. 2.59 D. 2.62 E. 2.65

Solution

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Let X1,X2,...,XnX_1, X_2, ... , X_n n observations from population with mean μ\mu and variance σ2\sigma^2. Then sample mean\textbf{sample mean} is defined by:

X=1ni=1nXi,\begin{align*} \overline{X}=\frac{1}{n}\sum_{i=1}^nX_i, \end{align*}

and it has normal distribution N(μ,σ2n)N(\mu, \frac{\sigma^2}{n}) and it is point estimate of μ\mu, e.c. μ^=X\widehat{\mu}=\overline{X}.

Standard error\textbf{Standard error} of x\overline{x} is σn\frac{\sigma}{\sqrt{n}}. But in many cases standard deviation is unknown and we can replace it with sample standard deviation s, where s=1n1i=1n(XiX)2s=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline{X})^2 is sample standard deviation.

The sample is 7, 9, 14, 15, 22.

x=7+9+14+15+225=13.4,s=(713.4)2+(913.4)2+(1413.4)2+(1513.4)2+(2213.4)251=137.24=34.3=5.86,Standard error(x)=5.865=2.62.\begin{align*} \overline{x}&=\frac{7+9+14+15+22}{5}=13.4,\\ s&=\sqrt{\frac{(7-13.4)^2+(9-13.4)^2+(14-13.4)^2+(15-13.4)^2+(22-13.4)^2}{5-1}}\\ &=\sqrt{\frac{137.2}{4}}=\sqrt{34.3}=5.86,\\ \text{Standard error}(\overline{x})&=\frac{5.86}{\sqrt{5}}=\fbox{2.62}. \end{align*}

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