Answer :
Answer:
Hypothesis Test states that we will accept 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] = 8.54 and [tex]X_2bar[/tex] = 8.69
[tex]s_1[/tex] = 0.225 and [tex]s_2[/tex] = 0.725
[tex]n_1[/tex] = 37 and [tex]n_2[/tex] = 58
[tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }[/tex] = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.
Test statistics = [tex]\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }[/tex] follows [tex]t_9_3[/tex]
= -1.219
At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that mean voltage for these two types of batteries is same.