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Sagot :
Answer:
a. difference of means
Step-by-step explanation:
Given that :
Mean , x = 9.4
Standard deviation, [tex]s.d_1[/tex] = 2.5
Number, [tex]n_1[/tex] = 12
Mean, y = 6.5
standard deviation, [tex]s.d_2[/tex] = 2.4
Number, [tex]n_2[/tex] = 12
The null hypothesis is : [tex]$H_0: \mu_1=\mu_2$[/tex]
The alternate hypothesis is : [tex]$H_1: \mu_1>\mu_2$[/tex]
Level of significance, [tex]\alpha[/tex] = 0.1
From the [tex]\text{standard normal table, right tailed,}[/tex] [tex]$t_{1/2}$[/tex] = 1.363
Since out test is right tailed.
Reject [tex]H_0[/tex], if [tex]$T_0>1.363$[/tex]
We use the test statics,
[tex]$t_0=\frac{(x-y)}{\sqrt{\frac{s.d_1}{n_1}+\frac{s.d_2}{n_2}}}$[/tex]
[tex]$t_0=\frac{(9.4-6.5)}{\sqrt{\frac{6.25}{12}+\frac{5.76}{12}}}$[/tex]
[tex]$t_0=2.899$[/tex]
[tex]$|t_0|=2.899$[/tex]
[tex]\text{Critical value}[/tex]
The value of [tex]$|t_{1/2}|$[/tex] with minimum [tex]$\left(n_1-1,n_2-1)$[/tex] that is 11 df is 1.363
We go [tex]$|t_0|=2.899$[/tex] and [tex]$|t_{1/2}|$[/tex] = 1.363
Decision making:
Since the value of [tex]|t_0|>|t_{1/2}|$[/tex] and we reject the [tex]H_0[/tex]
The p-value : right tail [tex]H_a:(p>2.8988)[/tex]
= 0.00724
Therefore the value of [tex]$p_{0.1} > 0.00724$[/tex], and so we reject the [tex]H_0[/tex]
Thus we are testing 'the difference of means" in this problem.
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