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inysia [295]
3 years ago
6

Please help me with this NO LINKS OR BOT ANSWERS PLEASE

Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

You realize the "bots" ANSWER THE QUESTIONS AUTOMATICALLY. IF they are bots ofc AND by capitalizing your words, IT MAKES IT MORE LIKELY TO SEE THEM!!! AND PEOPLE WANT TO ANNOY IDIOTS WHO PUT THAT IN THEIR QUESTION

Step-by-step explanation:

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A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

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

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

where, \bar X = sample mean weight = 31.8 ounces

             s = sample standard deviation = 0.48 ounces

             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0
4 years ago
The results of a poll show that the percent of people in a town who want an observatory built on a nearby mountain is in the int
V125BC [204]
In finding this value you average lower and upper bound
  (0.6+0.82)/2 = 0.71   =0.71  estimated    margin of error = distance from  estimate point lower/ upper bound   This interval will be twice margin error
   (0.82-0.6)/2 = 0.11    How far is 0.82 from 0.71 ??   =0.11=11%

7 0
4 years ago
ASAP! <br> may someone help me? It’s a test!
wel

Answer:

A

Step-by-step explanation:

5 0
3 years ago
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CAN someone help me PLZZZZ<br><br>Find the rate change of rom 2008 to 2016?
anyanavicka [17]

Answer:

22.25

Step-by-step explanation:

1302-1124/2016-2008

6 0
3 years ago
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