27. A manufacturer claims its Brand A battery lasts longer than its competitor's Brand B battery. Nine batteries of

Mathematics

1 answer:

mamaluj [8]2 years ago

4 0

According to the desired test, the test that should be conducted is:

(B) A one-sided, two-sample t-test

At the null hypothesis, it is tested if Brand A batteries do not last longer than Brand B batteries, that is, the result of the subtraction is of at most 0:

$H_0: \mu_A - \mu_B \leq 0$

At the alternative hypothesis, it is tested if it is greater, that is:

$H_1: \mu_A - \mu_B > 0$

The factors determining the test used are as follows:

We can find the <u>standard deviation for the sample</u>, hence a t-test is used.

There are two samples, the lifetimes for batteries A and for batteries B, hence, a two-sample test is used.

We are testing if one mean is greater than another, and more than/less than tests are one-sided.

Hence, option B is correct.

You can learn more about test hypothesis at brainly.com/question/13873630

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Answer:

Probability that this whole shipment will be accepted is 0.7324.

Step-by-step explanation:

We are given that a company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 30 components

r = number of success = fewer than 3

p = probability of success which in our question is % rate

of defects, i.e; 6%

<em>LET X = Number of defective components</em>

So, it means X ~ $Binom(n=30, p=0.06)$

Now, Probability that this whole shipment will be accepted only when there are fewer than 3 defectives = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{30}{0}\times 0.06^{0} \times (1-0.06)^{30-0}+ \binom{30}{1}\times 0.06^{1} \times (1-0.06)^{30-1}+ \binom{30}{2}\times 0.06^{2} \times (1-0.06)^{30-2}$