Last time even if u have to put it in the comments <br> Anybody else wants to hangout me

Listed below are time intervals (min) between eruptions of a geyser. Assume that the "recent" times are within the past few y

Shalnov [3]

Answer:

p value = 0.039

t = - 2.169

Step-by-step explanation:

Applying the null and alternate hypothesis

$H_{0} : U1 = U2$

$H_{\alpha } : U1 \neq U2$

using excel worksheet to calculate for ( t and p )

t = -2.169

p = 0.039

from the results obtained

The conclusion is affected by the significance level because : 0.1 < p > 0.01

so when the significance level is = 0.1 the Null hypothesis is rejected and we can say the mean time interval will change while

if the significance level = 0.01 the Null hypothesis is accepted and we can not say the mean time interval has changed because the p -value is greater than 0.01

attached is the excel solution

7 0

3 years ago

270 peaches are placed into three groups: small peaches,

Ghella [55]

Answer:

$0.45

Step-by-step explanation:

ratio is 2:4:3

total is 9

270 ÷ 9 = 30

smoll

60 × 0.25 = $15

med

120 × 0.30 = $36

Big

91.50 - 15 - 36 = 40.5

40.5 ÷ 90 = $0.45

4 0

3 years ago

Paul biked 55 miles in 10 hours . what is the unit rate of miles per hour

irinina [24]

Answer:

5.5 miles per hour

Step-by-step explanation:

Since he travels a total of 55 miles in 10 hours, you need to divide.

55/10 = 5.5

total hours/total miles = miles per hour

3 0

3 years ago

Read 2 more answers

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0

3 years ago

Which of the following rational functions is graphed below?

GREYUIT [131]

Answer:

Answer D.

Step-by-step explanation:

After plugging in each function, option D's graph is identical to the included image.

5 0

4 years ago

Last time even if u have to put it in the comments Anybody else wants to hangout me