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Marat540 [252]
3 years ago
9

Last time. (fifty each) F ree p oints! Jess is signing out, but want to make everyone's day. Have a great day!

Mathematics
2 answers:
Svetllana [295]3 years ago
6 0

Answer:

Umm.. thanks so much

Step-by-step explanation:

gulaghasi [49]3 years ago
6 0

Answer:

well then

Step-by-step explanation:

i shall take these points with gratitude.

i thank you with the power of god and anime

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y=0x-2

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A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

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mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁:  μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}  } }

Z≈N(0;1)

Z= \frac{(533-340)-120}{\sqrt{\frac{84^2}{56} +\frac{93^2}{102}  } }= 5.028

I hope this helps!

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