Mathematics: An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77 type Q batteries. The mean voltage is measured as 9.29 for the type K batteries with a standard deviation of 0.374, and the mean voltage is 9.65 for type Q batteries with a standard deviation of 0.518. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1 be the true mean voltage for type K batteries and

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An engineer is comparing voltages for two types

Ответ:

lashondrastokes83

18.01.2022

Mathematics

1. Null Hypothesis, H_0 : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, H_1 : $\mu_1\neq \mu_2$ {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, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, H_0 : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, H_1 : $\mu_1\neq \mu_2$ {mean voltage for these two types of

batteries is different]

The test statistics we use here will be :

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows t_n__1+n_2-2

where, X_1bar = 9.29 and X_2bar = 9.65

s_1 = 0.374 and s_2 = 0.518

n_1 = 83 and n_2 = 77

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2} }{83+77-2} }$ = 0.45 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77} } }$ follows t_1_5_8

= -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.