I need help with number 5 ASAPPPPP!!!!!!

A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of

finlep [7]

Answer:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

$p_v =2*P(t_{(7)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:

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

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=0.475$

The sample deviation calculated $s=0.183$

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 0.5$

Alternative hypothesis: $\mu \neq 0.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=8-1=7$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(7)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

4 0

3 years ago

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What is the slope of the line that contains these points?

seraphim [82]

Answer:

8.

Step-by-step explanation:

The slope =

difference in y coordinates of 2 points / difference in coordinates of corresponding x coordinates.

So taking the first 2 points:

The slope = (18-10) / 0 - (-1)

= 8/1

= 8.

This is confirmed by slope between the second and third points

slope = 26-18/ (1-0) = 8.

8 0

3 years ago

Alice’s backyard is a rectangular piece of property that is twice as long as it is wide. The total area of her yard is 1,000 m2.

irina [24]

The approximate width of Alice's backyard is 22.36 in.

Let l be the length of the rectangle and w be the width.
l = 2w
A = 1000 m2 = l*w
1000 m2 = 2w *w
500 m2 = w^2
√500 m2 = √w^2
w = <span>22.360679775 in or 22.36 in</span>

Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.

8 0

3 years ago

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If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, a

Karolina [17]

Let h = height of the box,
x = side length of the base.

Volume of the box is $V=x^{2} h = 115$ .
So $h = \frac{115}{ x^{2} }$

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
$S = 2( x^{2} + hx + hx) \\ = 2 x^{2} + 4hx \\ = 2 x^{2} + 4( \frac{115}{ x^{2} } )x$
$= 2 x^{2} + \frac{460}{x}$
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

$S' = 4x- \frac{460}{ x^{2} } = 0$
$4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86$
Then $h= \frac{115}{4.86^{2}} = 4.87$
So the box is 4.86 in. wide and 4.87 in. high.

5 0

3 years ago

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Which shows the expressions in the order they would appear on a number line from least to greatest?

love history [14]

$\frac{17}{9} =1.888888889
$
$\sqrt{6} =2.449489743$
$\sqrt{30}=5.477225575$
$3^{3}=27$

6 0

3 years ago

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