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Sonbull [250]
3 years ago
12

April lost 5 pounds in her first week of college. Over the next three weeks she lost 3 pounds, gained 2 pounds, and then lost 1

pound. Over the four weeks, what was the change in April's weight? Give a single integer to represent your answer (don't forget to include a negative sign if applicable).
Mathematics
1 answer:
Doss [256]3 years ago
4 0

Change in April's weight is -7

<em><u>Solution:</u></em>

Let the initial weight be "x"

April lost 5 pounds in her first week of college

First week = lost 5 pounds

Therefore, x - 5

Over the next three weeks she lost 3 pounds, gained 2 pounds, and then lost 1 pound

<em><u>Next three week:</u></em>

Lost 3 pound

Gained 2 pound

Lost 1 pound

<em><u>Therefore, weight after 4 weeks is given as:</u></em>

x - 5 - 3 + 2 - 1 = x -7

After 4 week, the weight is x - 7

<em><u>The change in April weight is given as:</u></em>

Weight after 4 week - initial weight = x - 7 - x = -7

Thus change in April's weight is -7 (Negative sign represents loss in weight )

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Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let \mu = <u><em>population mean length of the board</em></u>.

So, Null Hypothesis, H_0 : \mu = 2564.0 millimeters    {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, H_A : \mu \neq 2564.0 millimeters      {means that the boards are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                             T.S.  =  \frac{\bar X-\mu}{\frac{s }{\sqrt{n}} }  ~  t_n_-_1

where, \bar X = sample mean length of boards = 2559.5 millimeters

            s = sample standard deviation = 15.0 millimeters

             n = sample of boards = 26

So, <em><u>the test statistics</u></em> =  \frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }  ~   t_2_5

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The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

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