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navik [9.2K]
3 years ago
13

An automobile manufacturer claims that its car has a 37.2 miles/gallon (MPG) rating. An independent testing firm has been contra

cted to test the MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 120 cars, they found a mean MPG of 37.1. Assume the standard deviation is known to be 1.1. A level of significance of 0.05 will be used. Make a decision to reject or fail to reject the null hypothesis.
Mathematics
1 answer:
tatiyna3 years ago
6 0

Answer:

fail to reject the null hypothesis

Step-by-step explanation:

Based on the information provided I would fail to reject the null hypothesis. This is because the Miles per gallon that the independent testing firm has concluded for the car nearly matches the car's MPG claim. The difference being 0.1 . Since the standard deviation is 1.1 for this test, this means that the average falls perfectly within this range and makes the data very compelling. Combine this with a very low level of significance, I would say that the claim is backed by the data.

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Step-by-step explanation:

<u>Given Equation is:</u>

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The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an
solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. ​(a) What is the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information X\sim N(\mu=268, \sigma^{2}=12^{2}).

(a)

The minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

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Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283

Thus, the minimum pregnancy length that can be in the top 11​% of pregnancy​ lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248

Thus, the maximum pregnancy length that can be in the bottom ​5% of pregnancy​ lengths is 248 days.

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Answer:

2+5=7

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