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

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: