Paired t‐Test for Mean Comparison with Dependent Samples To study the effects of an advertising campaign at a supply chain, several stores are randomly selected with the following observed before‐ and after‐advertising monthly sales revenues: Store number 1 2 3 4 5
Old sales revenue (mil. $) 5.2 6.5 7.2 5.7 7.6
New sales revenue (mil. $) 6.4 7.8 6.8 6.5 8.2
Let μ₁ and μ₂ be the means of old and new sales revenues, both in millions of dollars per month.
(a) At α = 0.05, test H0: μ2 ≤ μ1 versus H1: μ2 > μ1. Sketch the test. Interpret your result.
(b)Sketch and find the p‐value of the test. Would you reject H0 if α = 0.01?

Answers

Answer 1

Answer:

Answer:

a) $t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308$

$p_v =P(t_((4))>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

b) The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

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.

Let put some notation

1=test value old , 2 = test value new

1: 5.2 6.5 7.2 5.7 7.6

2: 6.4 7.8 6.8 6.5 8.2

The system of hypothesis for this case are:

Null hypothesis: $\mu_2- \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 >0$

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

d: 1.2, 1.3, -0.4, 0.8, 0.6

The second step is calculate the mean difference

$\bar d= (\sum_(i=1)^n d_i)/(n)=0.7$

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

$s_d =(\sum_(i=1)^n (d_i -\bar d)^2)/(n-1) =0.678$

The 4 step is calculate the statistic given by :

$t=(\bar d -0)/((s_d)/(√(n)))=(0.7 -0)/((0.678)/(√(5)))=2.308$

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

$df=n-1=5-1=4$

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

$p_v =P(t_((4))>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

Part b

The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.

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Select the correct answer. Simplify the following expression. x-2/3 times x 6/7

Answers

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Step-by-step explanation: I got it right

Answer:

4/7

Step-by-step explanation:

A circle has a radius of five an arc in the circle has a central angle of 162 what is the length of the arc enter an exact answer in terms of pie or use with 3.14 for pie and enter your answer as a decimal

Answers

Answer:

Length of arc = 14.13

Step-by-step explanation:

Given:

Radius of circle, r = 5

An arc in the circle has a central angle of 162°

To find the length of the arc, L, let's use the formula:

$L = 2\pi r (\theta)/(360)$

Where $\theta$ = 162°

Therefore, substituting figures, we have:

$L = 2*3.14*5 ((162)/(360))$

L = 31.4 * 0.45

L = 14.13

Length of arc is 14.13 units

A publisher reports that 42%42% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 250250 found that 35%35% of the readers owned a particular make of car. Find the value of the test statistic. Round your answer to two decimal places.

Answers

Answer:

$z=\frac{0.35 -0.42}{\sqrt{(0.42(1-0.42))/(250)}}=-2.24$

Step-by-step explanation:

Data given and notation

n=250 represent the random sample taken

$\hat p=0.35$ estimated proportion of readers owned a particular make of car

$p_o=0.42$ is the value that we want to test

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that that the percentage is actually different from the reported percentage.:

Null hypothesis: $p=0.42$

Alternative hypothesis: $p \neq 0.42$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .