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Brrunno [24]
3 years ago
13

5Y+8 -ly-8-8y-1"Your answer​

Mathematics
2 answers:
gulaghasi [49]3 years ago
7 0

Answer:

-4y-1

Step-by-step explanation:

(5y+-y+-8y)+(8+-8+-1)

-4y-1

sammy [17]3 years ago
3 0

Answer:

5y + 8 - 1y - 8 - 8y - 1

variables = -4y

numbers = -1

<h2><em><u>-4y - 1 is the answer</u></em></h2>
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A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
GREYUIT [131]

Answer:

(a) Null Hypothesis, H_0 : \sigma = 10 beats per minute  

     Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute  

(b) The value of chi-square test statistics is 35.704.

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(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute.

Let \sigma = <u><em>population standard deviation for the pulse rates of men</em></u>.

(a) So, Null Hypothesis, H_0 : \sigma = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is <u>One-sample chi-square test</u> for standard deviation;

                             T.S.  =  \frac{(n-1)\times s^{2} }{\sigma^{2} }  ~  \chi^{2}__n_-_1  

where, s = sample standard deviation = 10.1 beats per minute

            n = sample of men = 36

So, <u><em>the test statistics</em></u> =  \frac{(36-1)\times 10.1^{2} }{10^{2} }  ~ \chi^{2}__3_5

                                   =  35.70  4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

                    P-value = P(\chi^{2}__3_5 > 35.704) = <u>0.4360</u>

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

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Answer:

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Step-by-step explanation:

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