Mathematics, 06.05.2020 02:39 daverius3153

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5. The average weight is 29.0 and the standard deviation is 0.36. Assume the weight of the bags follows normal distribution. Does this provide strong evidence that the true weight is not 28.3 grams?

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Answer from: denarreiahspen5921

We conclude that the true mean weight is not 28.3 grams at 5% significance level.

Step-by-step explanation:

We are given that some students checked 6 bags of Doritos marked with a net weight of 28.3 grams.

They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5. The average weight is 29.0 and the standard deviation is 0.36.

Let $\mu$ = true mean weight of bag.

So, Null Hypothesis, H_0 : $\mu$ = 28.3 grams {means that the true mean weight is 28.3 grams}

Alternate Hypothesis, H_A : $\mu$ $\neq$ 28.3 grams {means that the true mean weight is not 28.3 grams}

The test statistics that would be used here One-sample t test statistics as 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 weight = 29.0 grams

s = sample standard deviation = 0.36 grams

n = sample of bags = 6

So, test statistics = $\frac{29.0-28.3}{\frac{0.36}{\sqrt{6} } }$ ~ t_5

= 4.763

The value of t test statistics is 4.763.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -2.571 and 2.571 at 5 degree of freedom for two-tailed test.

Since our test statistics does not lie within the range of critical values of t, 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 the true mean weight is not 28.3 grams.

Answer from: Quest

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

Answer from: Quest

the required function is $y=-x+4$ .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-7}{)}=\frac{-3}{3}=-1$

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

$y=mx+b$

where, m is slope and b is y-intercept.

$y=-1x+4$

$y=-x+4$

therefore the required function is $y=-x+4$ . the graph of this function is shown below.

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Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weigh...

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