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4 years ago

In an experimental study, a lab wanted to divide volunteers into two

Mathematics

1 answer:

fredd [130]4 years ago

3 0

Step-by-step explanation:

Assigning the participants based on who shows up first would introduce bias. Those who arrive first may be more likely to already be punctual. Giving or not giving them the app may make it seem ineffective when it isn't.

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Ivanhoe Company is considering buying a new farm that it plans to operate for 10 years. The farm will require an initial investm

Ludmilka [50]

The NPV of this investment if the discount rate is 10 percent is: 1.58%.

<h3>Net present value (NPV)</h3>

Year Cash flow PVIF 10% Present value

0 ($11.86) 1.000 ($11.86)

1 1.90 0.909 $1.73

2 1.90 0.826 $1.57

3 1.90 0.751 $1.43

4 1.90 0.683 $1.30

5 1.90 0.621 $1.18

6 1.90 0.564 $1.07

7 1.90 0.513 $0.98

8 1.90 0.467 $0.89

9 1.90 0.424 $0.81

10 6.45 0.386 $2.49

NPV $1.58

1.9+5.25-2×35%=6.45

Hence, the NPV is $1.58.

Learn more about NPV here:brainly.com/question/17185385

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3 0

2 years ago

Explain how you can use doubles when multiplying with 4 to find 4 X 8

liq [111]

If you want to evaluate it is 4, 4*8 But if u want to solve the function operation then it is 4 * 8. 32

3 0

3 years ago

Compare 3.14 to 31.4 which is greater or least equal

expeople1 [14]

3.14 is smaller. the more the decimal is to the right the larger the number is

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!!

Gelneren [198K]

Answer:

The correct answer is 5

3 0

4 years ago

Read 2 more answers

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

3 years ago