Question 14 Marks
Read the following text and answer the following questions on the basis of the same:
A group of 7 weeks old chickens, reared on a high protein diet weight 12, 15, 11, 16, 14, 14, 16 ounces; a second group of 5 chickens, similarly treated except that they receive a low protein diet, weight 8, 10, 14, 10, 13 ounces. Given that the value of t for 10 degree of freedom at 5% level of significance is 2.23.
Q.1. Mean weight of old chickens, reared on a high protein diet is:
(A) 14 (B) 11 (C) 98 (D) 55
Q.2. Sum of the deviations of old chickens is:
(A) 23 (C) 22 (B) 18 (D) 14
Q.3. Mean weight chickens, receive a low protein diet is:
(A) 14 (B) 11 (C) 98 (D) 55
Q.4. The standard error for given two samples is:
(A) 2.14 (B) 2.41 (C) 2.93 (D) 2.55
A group of 7 weeks old chickens, reared on a high protein diet weight 12, 15, 11, 16, 14, 14, 16 ounces; a second group of 5 chickens, similarly treated except that they receive a low protein diet, weight 8, 10, 14, 10, 13 ounces. Given that the value of t for 10 degree of freedom at 5% level of significance is 2.23.
Q.1. Mean weight of old chickens, reared on a high protein diet is:
(A) 14 (B) 11 (C) 98 (D) 55
Q.2. Sum of the deviations of old chickens is:
(A) 23 (C) 22 (B) 18 (D) 14
Q.3. Mean weight chickens, receive a low protein diet is:
(A) 14 (B) 11 (C) 98 (D) 55
Q.4. The standard error for given two samples is:
(A) 2.14 (B) 2.41 (C) 2.93 (D) 2.55
Answer
View full question & answer→(1) (A) 14
Explanation: Number of terms in first group i.e. $n_1$=7
Mean weight of old chickens, reared on a high protein diet is
$\begin{aligned} \bar{x} & =\frac{\sum x}{n_1} \\ & =\frac{12+15+11+16+14+14+16}{7} \\ & =\frac{98}{7}=14\end{aligned}$
(2) (C) 22
Explanation: Sum of the squares of mean deviations of old chickens is
$\sum(x-\bar{x})^2=\left(x_1-\bar{x}\right)^2+\left(x_2-\bar{x}\right)^2+\ldots+\left(x_y-\bar{x}\right)^2$
$=(12-14)^2+(15-14)^2+(11-14)^2$$+(16-14)^2+(14-14)^2+(14-14)^2$$+(16-14)^2$
= 4 + 1 + 9 + 4 + 0 + 0 + 4
= 22
(3) (B) 11
Explanation: Number of terms in second group i.e. $n_1=5$
Mean weight chickens, receive a low protein diet is
$\begin{aligned} \bar{y} & =\frac{y}{n_1} \\ & =\frac{8+10+14+10+13}{5} \\ & =\frac{55}{5}=11\end{aligned}$
(4) (A) 2.14
Explanation: Formula for standard error is
$s=\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}}$
We already have the following values
Now, $\sum(y-\bar{y})^2=\left(y_1-\bar{y}\right)^2+\left(y_2-\bar{y}\right)^2+\ldots$$+\left(y_7-\bar{y}\right)^2$
$=(8-11)^2+(10-11)^2+(14-11)^2$$+(10-11)^2+(13-11)^2$
= 9 + 1 + 9 + 1 + 4
= 24
So,$s=\sqrt{\frac{22+24}{7+5-2}}=\sqrt{\frac{46}{10}}$
$=\sqrt{4.6}=2.14$
Explanation: Number of terms in first group i.e. $n_1$=7
Mean weight of old chickens, reared on a high protein diet is
$\begin{aligned} \bar{x} & =\frac{\sum x}{n_1} \\ & =\frac{12+15+11+16+14+14+16}{7} \\ & =\frac{98}{7}=14\end{aligned}$
(2) (C) 22
Explanation: Sum of the squares of mean deviations of old chickens is
$\sum(x-\bar{x})^2=\left(x_1-\bar{x}\right)^2+\left(x_2-\bar{x}\right)^2+\ldots+\left(x_y-\bar{x}\right)^2$
$=(12-14)^2+(15-14)^2+(11-14)^2$$+(16-14)^2+(14-14)^2+(14-14)^2$$+(16-14)^2$
= 4 + 1 + 9 + 4 + 0 + 0 + 4
= 22
(3) (B) 11
Explanation: Number of terms in second group i.e. $n_1=5$
Mean weight chickens, receive a low protein diet is
$\begin{aligned} \bar{y} & =\frac{y}{n_1} \\ & =\frac{8+10+14+10+13}{5} \\ & =\frac{55}{5}=11\end{aligned}$
(4) (A) 2.14
Explanation: Formula for standard error is
$s=\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}}$
We already have the following values
Now, $\sum(y-\bar{y})^2=\left(y_1-\bar{y}\right)^2+\left(y_2-\bar{y}\right)^2+\ldots$$+\left(y_7-\bar{y}\right)^2$
$=(8-11)^2+(10-11)^2+(14-11)^2$$+(10-11)^2+(13-11)^2$
= 9 + 1 + 9 + 1 + 4
= 24
So,$s=\sqrt{\frac{22+24}{7+5-2}}=\sqrt{\frac{46}{10}}$
$=\sqrt{4.6}=2.14$
