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3 years ago

How do you do this?Can you help me with the steps? I don't really get this a lot

Mathematics

2 answers:

Tju [1.3M]3 years ago

8 0

Think of it as dividing fractions

polet [3.4K]3 years ago

4 0

Before you do the steps, change the fractions given to improper fractions:
2 2/3 = 8/3
3 1/6 = 19/6

Now, I think the first step they want you to write is:
8/3 (divided by) 19/6

When you divide two fractions, it's the same as multiplying the first fraction by the reciprocal of the second fraction. (A reciprocal is a fancy way of saying you flip the fraction)
So the next step will look like this:
8/3 x 6/19

With this setup, you can cross simplify the fractions so that they look like this:
8/1 x 2/19
When you multiply this, you get 16/19
Let me know if you have any questions!

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Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

Let $\mu$ = average miles for deluxe tires

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, test statistics = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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