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choli [55]
3 years ago
14

Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t

he hypothesis that the average age of a Zappos employee is less than 36 years old. A random sample of 22 Zappos employees was found to have an average age of 33.9 years. The standard deviation for this sample was 4.1 years. Amazon would like to set α = 0.025. The conclusion for this hypothesis test would be that because the test statistic is
Mathematics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

df = n-1= 22-1=21

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 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)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:  

Null hypothesis:\mu \geq 36  

Alternative hypothesis:\mu < 36  

The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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solniwko [45]

Answer:

1000000000000000000001

Step-by-step explanation:

Done. You're welcome

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3 years ago
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There are 790 cal in 610 ounce bottles of apple juice how many calories are there in one 10 ounce bottle of apple juice
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alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

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Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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Step-by-step explanation:

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