Respuesta :
Answer:
1. Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 = \mu_2[/tex] {mean voltage for these two types of
batteries is same}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_1\neq \mu_2[/tex] {mean voltage for these two types of
batteries is different]
2. Test Statistics value = -5.06
4. Decision for the hypothesis test is that we will reject null hypothesis.
Step-by-step explanation:
We are given that an engineer is comparing voltages for two types of batteries (K and Q).
where, [tex]\mu_1[/tex] = true mean voltage for type K batteries.
[tex]\mu_2[/tex] = true mean voltage for type Q batteries.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 = \mu_2[/tex] {mean voltage for these two types of
batteries is same}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_1\neq \mu_2[/tex] {mean voltage for these two types of
batteries is different]
The test statistics we use here will be :
[tex]\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows [tex]t_n__1+n_2-2[/tex]
where, [tex]X_1bar[/tex] = 9.29 and [tex]X_2bar[/tex] = 9.65
[tex]s_1[/tex] = 0.374 and [tex]s_2[/tex] = 0.518
[tex]n_1[/tex] = 83 and [tex]n_2[/tex] = 77
[tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2} }{83+77-2} }[/tex] = 0.45 Here, we use t test statistics because we know nothing about population standard deviations.
Test statistics = [tex]\frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77} } }[/tex] follows [tex]t_1_5_8[/tex]
= -5.06
At 0.05 or 5% level of significance t table gives a critical value between (-1.98,-1.96) to (1.98,1.96) at 158 degree of freedom. Since our test statistics is less than the critical table value of t as -5.06 < (-1.98,-1.96) so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that mean voltage for these two types of batteries is different.
