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Lemur [1.5K]
3 years ago
8

Can you please make a graph and submit it? Thanks!

Mathematics
1 answer:
rosijanka [135]3 years ago
6 0
(math)(way)(.com) hope it helps! sorry about the parentheses the site wouldn't let me type all together

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The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0
3 years ago
Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun
Evgen [1.6K]
The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.

95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
5 0
3 years ago
3x+2y=-16<br> -3x-8y=46<br> Find the solution of the system of equations
Mariulka [41]
For the first one Y = -6

3x + 2Y = -16
-2y -2y
3x = -18y
— —
3x 3x

Then divide -18y divided by 3X
To get Y= -6

Hope that helped!
5 0
3 years ago
A circular running track is 1/4 mile long. Brie runs on this track, completing each lap in 1/20 of an hour. What is Brie's runni
Minchanka [31]

Answer:

number 3

Step-by-step explanation:

6 0
2 years ago
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The units digit of a two digit number is 4 times the tens digit. If the digits are reversed, the new number is 54 more than the
Ratling [72]
The numbers are 82 and 28. 
5 0
3 years ago
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