Mathematics, 12.02.2020 03:28, Evycin

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 μ2 be the true mean voltage for type Q batteries. Use a 0.05 level of significance.
1. state the null and alternative hypothesis for the test
2. Compute the value of the test statistic. Round your answer to two decimal places
3. Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places
4. Make the decision for the hypothesis test a) reject null hypothesis b) fail to reject null hypothesis

Answers: 2

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answered: AreYouMyNewDad

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.

answered: Guest

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]

answered: Guest

step-by-step explanation:

only stephen hawking would know.