15,18,16,19,14,12,14,12,17,15,11
A) find the sample mean and standard deviation
B) state your null hypothesis and your alternate hypothesis
+33665 To find the sample mean add them all up and divide by how many there are:
$${\mathtt{mean}} = {\frac{\left({\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{17}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}}\right)}{{\mathtt{11}}}} \Rightarrow {\mathtt{mean}} = {\mathtt{14.818\: \!181\: \!818\: \!181\: \!818\: \!2}}$$
To find the sample standard deviation use the following formula:
$$sd = \sqrt{\frac{\sum_{i=1}^n(x_i-mean)^2}{n-1}}$$
where n is the number of items (11 here), xi are the individual data values (15, 18 ... etc.), and $$\sum_{i=1}^n$$ means sum all of the squared bracketed terms that follow.
I'll leave you to plug in the numbers.
You give no information about what comparison is to be done, so there is no way I can say anything about what the null hypothesis and its alternate might be!
+33665 To find the sample mean add them all up and divide by how many there are:
$${\mathtt{mean}} = {\frac{\left({\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{17}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}}\right)}{{\mathtt{11}}}} \Rightarrow {\mathtt{mean}} = {\mathtt{14.818\: \!181\: \!818\: \!181\: \!818\: \!2}}$$
To find the sample standard deviation use the following formula:
$$sd = \sqrt{\frac{\sum_{i=1}^n(x_i-mean)^2}{n-1}}$$
where n is the number of items (11 here), xi are the individual data values (15, 18 ... etc.), and $$\sum_{i=1}^n$$ means sum all of the squared bracketed terms that follow.
I'll leave you to plug in the numbers.
You give no information about what comparison is to be done, so there is no way I can say anything about what the null hypothesis and its alternate might be!
