Starla is measuring the floor of a rectangular room to determine how many square feet of a tile to buy to cover the floor

A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa

sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) $p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference $d_i=A_i-B_i$ and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_A- \mu_B = 0$

Alternative hypothesis: $\mu_A -\mu_B \neq 0$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0

2 years ago

Leena walked 2/3 of a mile What is 2/3 written as a sum of unit fractions with the denominator of nine

aliya0001 [1]

Bear in mind that multiplying anything, and anything whatsoever by 1, has a product of the anything itself.

so, let's multiply 2/3 by 1 then and check about,

$\bf \cfrac{same}{same}=1\qquad thus\cfrac{2}{3}\cdot 1\implies \cfrac{2}{3}\cdot \cfrac{3}{3}\implies \cfrac{6}{9}
\\\\\\
\textit{now, we can split that and write it as a sum}\implies \cfrac{1+5}{9}\implies \cfrac{1}{9}+\cfrac{5}{9}$

3 0

2 years ago

Ken and Hamid run around a track. It takes Ken 80 seconds to complete a lap. It takes Hamid 60 seconds to complete a lap. Ken an

likoan [24]

The <em><u>correct answer</u></em> is:

Ken will have run 3 laps and Hamid will have run 4.

Explanation:

To find this, we first find the number of seconds that will have passed when they meet again. We use the LCM, or least common multiple, for this. First we find the prime factorization of each number:

80 = 10(8)

10 = 5(2)

8 = 2(4)

4 = 2(2)

80 = 2(2)(2)(5)(2)

60 = 10(6)

10 = 5(2)

6 = 2(3)

60 = 2(2)(3)(5)

For the LCM, we multiply the common factors by the uncommon. Between the two numbers, the common factors are 2, 2 and 5. This makes the uncommon 2, 2, and 3, and makes our LCM

2(2)(5)(2)(2)(3) = 240

This means every 240 seconds they will both be at the start line.

Since Ken completes a lap in 80 seconds, he completes 240/80 = 3 laps in 240 seconds.

Since Hamid completes a lap in 60 seconds, he completes 240/60 = 4 laps in 240 seconds.

7 0

1 year ago

Read 2 more answers

Frank started out in his car travelling 45 mph. When Frank was 1 3 miles away, Daniel started out from the same point at 50 mph

oee [108]

Answer:

It will take 2.6 hours by Daniel to catch up with Frank.

Step-by-step explanation:

Let Daniel will catch up with Frank after t hours.

Now, Frank will go 13 miles in $\frac{13}{45} = 0.288$ hours.

So, the distance traveled by Frank in (t + 0.228) hours with 45 mph speed will be equal to the distance traveled by Daniel in t hours with 50 mph speed.

Hence, 45(t + 0.288) = 50t

⇒ 5t = 13

⇒ t = 2.6 hours.

Therefore, it will take 2.6 hours by Daniel to catch up with Frank. (Answer)

8 0

2 years ago

The intersection of two planes can be a point.

gregori [183]

No

If two planes intersect each other, the intersection will always be a line. where r 0 r_0 r0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

4 0

1 year ago