Liz train for rock climbing 4 days a week

Mr.morales makes a table of the number of chaperones needed for different number of students attending a field trip. the ratio o

vichka [17]

Answer:

mark brainest

Step-by-step explanation:

7 0

3 years ago

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Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist

satela [25.4K]

Answer:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

$p_v =2*P(t_{(9)}>1.539)=0.158$

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.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

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

And the sample deviation given by:

$s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And we got:

$\bar X=7.55$ represent the sample mean

$s=0.103$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05$ 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.

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 7.5$

Alternative hypothesis: $\mu \neq 7.5$

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:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

P-value

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

$df=n-1=10-1=9$

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

$p_v =2*P(t_{(9)}>1.539)=0.158$

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.

3 0

3 years ago

Help please I'm really confused I will mark a brainliest for correct answer

max2010maxim [7]

Easy do 12 times 16 divided by two = 96

answer=96

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4 years ago

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A certain ferry moves up and down a river between Town A and

Blizzard [7]

Answer: 60 miles

Step-by-step explanation:

Given

It takes ferry 2 hours to travel to town A and only $1.5\ hr$ to travel back to town B.

Speed of current is $u=5\ mph$

Suppose the speed of the ferry is $v$ mph

Distance traveled in both the cases is same, but it took more time traveling to city A that is, ferry is moving upstream and downstream for returning time.

$\Rightarrow 2(v-5)=1.5(v+5)\\\Rightarrow 2v-10=1.5v+7.5\\\Rightarrow 0.5v=17.5\\\Rightarrow v=35\ mph$

Distance between the towns is $2\times (35-5)=60\ \text{miles}$

5 0

3 years ago

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please help me A portion of Taquan's check register is shown. Before Taquan made his utility payment on March 24, the balance of

Natalka [10]

The check register is used to show the total amount Taquan spent

Using the assumed values, Taquan's initial balance is $1336.26

<h3>How to determine the amount before making the utility payment?</h3>

The question is incomplete, as the check register is not shown.

So, I will give a general explanation

Assume that the utility bill is $241.46

From the question, the new balance on March 24 is $1094.80

This means that the initial balance is:

Initial balance = New balance + Utility bill

This gives

Initial balance = $1094.80 + $241.46

Evaluate the sum

Initial balance = $1336.26

Using the assumed values, Taquan's initial balance is $1336.26