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Answer:
Therefore we fail to reject the Null hypothesis [tex]H_0[/tex]
Because the is no sufficient evidence to support the claim of the alternative hypothesis [tex]H_a[/tex]
Step-by-step explanation:
From the question we are told that
Sample size [tex]n_1 =52[/tex]
Mean time 1 [tex]\=x_1=24.04\ hours[/tex]
Mean time 2 [tex]\=x_2=24.52\ hours[/tex]
Standard deviation [tex]\partial _1=5.769[/tex]
Standard deviation [tex]\partial _2=5.556[/tex]
Level of significance 0.05
The null hypothesis is [tex]H_o: \mu_1=\mu_2[/tex]
Alternative hypothesis [tex]H_a: \mu_1<\mu_2[/tex]
Generally the test statistics is mathematically represented as
[tex]Z= \frac{\=x_1-\=x_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}[/tex]
[tex]Z= \frac{24.04-24.52}{\sqrt{\frac{5.769^2}{52}+\frac{5.5562^2}{52}}}[/tex]
[tex]Z= -0.4321583379[/tex]
[tex]Z= -0.43[/tex]
From Table the Z-critical value at [tex]\alpha =0.05\ is\ -1.6449[/tex]
Therefore
Zcal> Zcritical
(-0.43)=(-1.645)
Generally we fail to reject the null hypothesis [tex]H_0[/tex] at [tex]\alpha = 0.05[/tex]
Therefore we fail to reject the[tex]H_0[/tex]
Because the is no sufficient evidence to support the claim of [tex]H_a[/tex]