Question

Six guinea pigs injected with 0.5 mg of a medication took on the average 15.4 seconds to fall asleep with an unbiased standard deviation of 2.2 seconds while six other guinea pigs injected with 1.5 mg of the medication took on the average 11.2 seconds to fall asleep with an unbiased standard deviation of 2.6 seconds. Use 5% level of significance to test the null hypothesis that the difference in dosage has no effect. ​

Answer 1

Answer:

answer will be the 5

Step-by-step explanation:

mark me brainliest

Answer 2

Answer:

Considering,

H₀ = the difference in the dosage has no kind of effect which means the two mean are equal.

H₁ = the difference in the dosage has any kind of effect which means the two means are not equal.

Now, the standard error of the difference in between two samples

(for small samples) where the variance in population is unknown.

Given :

2.2 =  σ₁ = s₁ x $\sqrt{n1(n1 - 1)}$

or, 2.2 = s₁ x $\sqrt{6/5}$ = s₁ x 1.095

or, s₁ = 2.2/1.095 = 2.009

Also, it is given that :

2.6 = σ₂ = s₂ x $\sqrt{n2(n2 - 1)}$

or, 2.6 = s₂ x 1.095

or. s₂  = 2.6/1.095  = 2.374

Now, Standard Error = $\sqrt{(s1^2 + s2^2)/(n - 1)}$ = $\sqrt \frac{(2.009)^2 + (2.374)^2}{6 - 1}$

or, Standard Error = 1.391

[here,  s₁ and s₂ are in biased conditions, so we divide it by n-1]

Now, t =  (Mean 1 - Mean 2) / Standard Error = (15.4 - 11.2)/1.391 = 3.01

therefore, at level 5% the value for 't' for 10 d.f is 2.228 which can be able to place at chance of the fluctuations but the value computed is greater than this data. so, we reject the H₀ and declare that the dosage is significant.

#SPJ2