91影视

The hypotheses of interest are

\({H_0}:\,\,\,{\mu _1} = {\mu _j},\,i \ne j\,\)

Versus alternative hypothesis

\({H_a}:\)At least two of the \({\mu _i}\)鈥榮 is different,

Where is the true average survival time of corresponding regimen.

Notice first that

\(I = 6\),treatments,

and

\(J = 14,\)samples of each type 鈥� rats

The following table needs to be filled with corresponding values:

The degree of freedom is

\(\begin{aligned}{l}\,\,\,\,\,\,\,\,\,\,I - 1 = 6 - 1 = 5;\\I \cdot \left( {J - 1} \right) = 6 \cdot \left( {14 - 1} \right) = 78;\\\,\,\,\,\,\,\,\,I \cdot J - 1 = 6\, \cdot `14 - 1 = 83.\end{aligned}\)

Denote with

\(\begin{aligned}{l}{x_i} = \sum\limits_{j = 1}^J {{x_{i\,j}}} \,;\\{x_{ \cdot \cdot }} = \sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J {\,{x_{i\,j}}} \,;\,} \end{aligned}\)

The

total sum of squares

(SST),

treatment sum of squares

(SSTr), and

error sum of squares

(SSE) are given by

\(\begin{aligned}{l}\,\,\,\,\,SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J \, \left( {\,{x_{i\,j}} - \mathop x\limits^\_ ..} \right)} ^2} = \sum\limits_{i = 1}^I \, \sum\limits_{j = 1}^J \, x_{_{i\,j}}^2 - \frac{1}{{I - J}}x_{ \cdot \cdot }^2\,;\\SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J \, \left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = \frac{1}{J} \cdot \sum\limits_{i = 1}^I \, x_{_{i\,.}}^2 - \frac{1}{{I - J}}x_{ \cdot \cdot }^2\,;\\\,\,\,\,\,SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J \, \left( {\,{x_{i\,j}} - {{\mathop x\limits^\_ }_{i.}}} \right)} ^2} = \left( {J - 1} \right)\sum\limits_{i = 1}^i \, s_i^2\end{aligned}\)

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.\,SSTr;\\\,\,\,\,\,MSE = \frac{1}{{I\,\, \cdot \,\,\left( {J - 1} \right)}}.\,SSE\,.\end{aligned}\)

F is ratio of the two mean squares

\(F = \frac{{MSTr}}{{MSE}}\)

Compute all the values one by one. Values of

\({\mathop x\limits^\_ _{i.}}\, = \frac{1}{J} \cdot {x_{i\,.}}\)

Are given in the exercise as

\(\begin{aligned}{l}{\mathop x\limits^\_ _{1.}}\, = 166;\\{\mathop x\limits^\_ _{2.}}\, = 303;\\{\mathop x\limits^\_ _{3.}}\, = 266;\\{\mathop x\limits^\_ _{4.}}\, = 212;\\{\mathop x\limits^\_ _{5.}}\, = 202;\\{\mathop x\limits^\_ _{6.}}\, = 184;\end{aligned}\)

The grand mean \(\mathop x\limits^\_ ..\) is

\(\begin{aligned}{l}\mathop x\limits^\_ .. = \frac{1}{{I - J}}\,\,\sum\limits_{i = 1}^I \, \sum\limits_{j = 1}^J \, {x_{i\,j}} = \frac{1}{I}.\,\sum\limits_{i = 1}^6 \, {\mathop x\limits^\_ _{i.}}\,\\ = \frac{1}{6}.\left( {166 + 303 + 266 + 212 + 202 + 184} \right)\\ = 222.17\end{aligned}\)

The treatment sum of squares is

\(\begin{aligned}{l}SSTr = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J \, \left( {{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = J\, \cdot \,\,\sum\limits_{i = 1}^6 \, \cdot {\left( {{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\, = 14.\left( {{{\left( {166 - 222.17} \right)}^2} + {{\left( {303 - 222.17} \right)}^2} + \cdots + {{\left( {184 - 222.17} \right)}^2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = 14.13,\,577.83\\\,\,\,\,\,\,\,\,\,\,\,\, = 190,\,076\end{aligned}\)

The error sum of squares is

\(\begin{aligned}{l}SSE = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^J \, \left( {\,{x_{i\,j}} - {{\mathop x\limits^\_ }_{i.}}} \right)} ^2} = \left( {J - 1} \right)\sum\limits_{i = 1}^i \, s_i^2\\\,\,\,\,\,\,\,\,\,\,\,\, = \left( {14 - 1} \right).\,\left( {{{32}^2} + {{53}^2} + \cdots + {{31}^2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = 13.10,\,091\\\,\,\,\,\,\,\,\,\,\,\,\, = 131,\,183\end{aligned}\)

Fundamental Identity:

\(SST = SSTr + SSE\).

Error sum of squares is

\(SST = SSTr + SSE = 190,\,076 + 131,\,183 = 321,\,\,259\)

The mean squares can be computed now as

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.\,SSTr = \frac{1}{5}.109,\,076 = 38,\,015.1;\\MSE = \frac{1}{{I\,\, \cdot \,\,\left( {J - 1} \right)}}.\,SSE = \frac{1}{{6\,\, \cdot \,\,\left( {14 - 1} \right)}} \cdot 131,\,183 = 1681.83\end{aligned}\)

The value of F statistic is

\(f = \frac{{MSTr}}{{MSE}} = \frac{{38,\,015.1}}{{1681.83}} = 22.6\)

The ANOVA table now becomes

As for usual tests, you can either make conclusion about the hypotheses look at the F critical value or a P value. Remember that the hypotheses of interest are

\({H_0}:\,\,\,{\mu _1} = {\mu _j},\,i \ne j\,\)

Versus alternative hypothesis

\({H_a}:\)At least two of the \(\mu - i\)鈥榮 is different,

The P value is the area to the right of f value under the F curve where F has Fisher's distribution with degrees of freedom 5 and 78; thus

\(P = P\left( {F > f} \right) = P\left( {F > 22.6} \right) \approx 0\)

Which was computed using software (you could estimate it using the table in the appendix). Because

\(P = 0 < 0.01 = \alpha \)

at given significance level. There is statistically significance difference in true averages among the six different survival times.

Using the table, you could use e.g., value \({F_{0.01,5,60}}\) (this is approximate) for which the area under the F curve to the right of \({F_{0.01,5,60}}\) is 0.01. The value is

\({F_{0.01,5,60}} = 4.76 < 22.6 = f\),

Which indicates to reject null hypothesis at significance level 0.01

First interval:

Mean square error is

\(MSE = 1681.83\)

Using the fact that random variable

\(\theta = {\mathop X\limits^\_ _1}\, - \,\frac{1}{5}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5} + {{\mathop X\limits^\_ }_6}} \right)\)

has approximately student distribution with \(\,\left( {J - 1} \right) = 78\) degrees of freedom, from the fact

\(P\left( {{t_{\alpha /2,78}}\, < \,{{\mathop X\limits^\_ }_1}\, - \,\frac{1}{5}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5} + {{\mathop X\limits^\_ }_6}} \right)\,\, < {t_{\alpha /2,78}}} \right) = 1 - \alpha \)

Where, in this case, \(\alpha = 0.02\) and from the table in the appendix is approximately

\({t_{\alpha /2,78}} = {t_{0.01,\,78}} = 2.645\)

The 98% confidence interval is \({\sigma ^2}\).

\(\sum\limits_{i = 1}^6 \, {c_i}{\mathop x\limits^\_ _i}. \pm {t_{\alpha /2,I(J - I\~)}}.\sqrt {\frac{{MSE\,\, \cdot \,\sum\nolimits_{i = 1}^6 {c_i^2} }}{J}\,} \)

Here MSE is estimate of variance \({\sigma ^2}\).

Since \({c_1} = 1\) and \({c_i} = - 1/5,\)\(i = 2,3, \ldots ,6\)

\(\sum\limits_{i = 1}^6 \, c_i^2 = 1.2\)

Also, estimate of \(\theta \) is

\(\theta = \sum\limits_{i = 1}^6 \, {c_i}{\mathop x\limits^\_ _i}. = {\mathop X\limits^\_ _1}\, - \,\frac{1}{5}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5} + {{\mathop X\limits^\_ }_6}} \right) = - 67.4\)

Where all values are given in the table in the exercise. Therefore, the 98% confidence interval for \(\theta \) is

\( - 67.4 \pm 2.645\, \cdot \,\sqrt {\frac{{1681.83 \cdot 1.2}}{{14}}} \)

or equally

\(\left( { - 99.16,\, - 35.64} \right)\)

Second interval:

Mean square error is

\(MSE = 1681.83\)

Using the fact that random variable

\(\theta = 0 \cdot {\mathop X\limits^\_ _1}\, + \,\frac{1}{4}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5}} \right)\,\, - {\mathop X\limits^\_ _6}\,\)

has approximately student distribution with \(\,\left( {J - 1} \right) = 78\) degrees of freedom, from the fact

\(P\left( {{t_{\alpha /2,78}}\, < \,{{\mathop {0 \cdot X}\limits^{\,\,\,\,\,\,\,\,\_} }_1}\, + \,\frac{1}{4}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5}} \right)\, - \,{{\mathop X\limits^\_ }_6} < {t_{\alpha /2,78}}} \right) = 1 - \alpha \)

Where, in this case, \(\alpha = 0.02\) and from the table in the appendix is approximately

\({t_{\alpha /2,78}} = {t_{0.01,\,78}} = 2.645\)

The 98% confidence interval is

\(\sum\limits_{i = 1}^6 \, {c_i}{\mathop x\limits^\_ _i}. \pm {t_{\alpha /2,I(J - I\~)}}.\sqrt {\frac{{MSE\,\, \cdot \,\sum\nolimits_{i = 1}^6 {c_i^2} }}{J}\,} \)

Here MSE is estimate of variance \({\sigma ^2}\).

Since \({c_1} = 0\),\({c_i} = - 1/4,\,\,i = 2,3, \ldots ,5\,\)and \({c_6} = - 1\)

\(\sum\limits_{i = 1}^6 \, c_i^2 = 1.25\)

Also, estimate of \(\theta \) is

\(\theta = \sum\limits_{i = 1}^6 \, {c_i}{\mathop x\limits^\_ _i}. = {\mathop {0 \cdot X}\limits^\_ _1}\, + \,\frac{1}{4}.\,\left( {{{\mathop X\limits^\_ }_2} + {{\mathop X\limits^\_ }_3} + {{\mathop X\limits^\_ }_4} + {{\mathop X\limits^\_ }_5}} \right) - \,{\mathop X\limits^\_ _6} = - 61.75\)

Where all values are given in the table in the exercise. Therefore, the 98% confidence interval for \(\theta \) is

\(61.75 \pm 2.645\, \cdot \,\sqrt {\frac{{1681.83 \cdot 1.25}}{{14}}} \)

or equally

\(\left( {29.34,\,\,\,94.16} \right)\)

Here, the result of part(a), part(b) is

1. Reject null hypothesis
2. \(\left( { - 99.16,\, - 35.64} \right)\), \(\left( {29.34,\,\,\,94.16} \right)\)