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Question 2 (4 points) In a study of cereal leaf beetle damage in oats, researcher measured...

Question 2 (4 points) In a study of cereal leaf beetle damage in oats, researcher measured the number of beetle larvae per st

Question 2 (4 points) In a study of cereal leaf beetle damage in oats, researcher measured the number of beetle larvae per stem in small plots of oats after randomly applying one of two treatments: no pesticide or malathion at the rate of 0.25 lb. per acre. Here are the summary statistics: n > Group 1 mg S Treatment Control Malathion 2 mg 121 43 3.47 1.36 1.21 0.52 Provide a 95% confidence interval for the difference in mean numbers of beetle larvae per stem for the two treatments. A) (1.844, 2.376) B) (-0.356, 0.851) C) (-1.569, 4.623) OD) (1.974, 2.246) OE) (0.150, 4.070)
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In a study of cereal leaf beetle in oats, researcher measured the number of beetle larvae per stem in small plot of oats after randomly applying one of two treatments : no pesticides or malathion at the rate of 0.25 lv. Per acre. Here the summary of Statistics is given by,

For Group 1 ( 1 mg ),

n_1 = 121. \bar x_1 = 3.47   s_1 = 1.21

For Group 2 ( 2 mg ),

n_2 = 43,   \bar x_2 = 1.36   s_2 = 0.52

TO Find : 95 % confidence interval for the difference in mean.

We know that ,

( 1- \alpha ) 100% confidence interval is given by,

((\bar x_1-\bar x_2)- Z_\alpha_/_2\sqrt ( \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}) ,  (\bar x_1-\bar x_2)+ Z_\alpha_/_2\sqrt ( \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}) )

Now ,

95 % confidence interval for difference in mean is,

( (3.47-1.36)- Z_0._0_5_/_2 \sqrt ( \frac{1.21^2}{121}+\frac{0.52^2}{43}) , (3.47-1.36)+Z_0._0_5_/_2 \sqrt ( \frac{1.21^2}{121}+\frac{0.52^2}{43}) )

= ( 2.11-Z_0._02_5 \sqrt ( \frac{1.4641}{121}+\frac{0.2704}{43}) , 2.11+Z_0._02_5 \sqrt ( \frac{1.4641}{121}+\frac{0.2704}{43}) )

=( 2.11-1.96 \sqrt (0.0121+0.00628) , 2.11+1.96 \sqrt (0.0121+0.00628) ). ......( z_0._0_2_5 = 1.96 by using z table )

= (  2.11-1.96 \sqrt (0.018388)  , 2.11+1.96 \sqrt (0.018388) )

= ( 2.11-1.96 * 0.1356 , 2.11+1.96 * 0.1356 )

= ( 2.11 - 0.26578 , 2.11+ 0.26578 )

= ( 1.8442 , 2.3757 )

i.e. 95 % confidence interval for the difference in mean number of beetle larvae per stem for the two treatments is ( 1.844 , 2.376 )

Answer : Option A is correct ( 1.844, 2.376 )

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