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saul85 [17]
3 years ago
14

You are thinking of employing a t procedure to test hypotheses about the mean of a population using a significance level of 0.05

. You suspect the distribution of the population is not Normal and may be moderately skewed. Which of the following statements is correct?
a. You should not use the t procedure because the population does not have a Normal distribution.
b. You may use the t procedure, but you should probably claim the significance level is only 0.10.
c. You may use the t procedure, provided your sample size is large, say, at least 40.
d. You may not use the t procedure because t procedures are robust to non-Normality for confidence intervals but not for tests of hypotheses.
Mathematics
1 answer:
ch4aika [34]3 years ago
7 0

Answer:

b. You may use the t procedure, but you should probably claim the significance level is only 0.10.

Step-by-step explanation:

A t-test is generally used to compare the mean of two groups when the data comes from a normal distribution. If the distribution is not normal we can still perform a t-test but the sample size should be large enough.

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Josh invested $12,000 in mutual funds and received a sum of $20,000 at the end of the investment period. Calculate the ROI.
tresset_1 [31]

Answer:

ROI = 66.67%

Step-by-step explanation:

Given:

josh investment= $12,000

Received sum= $20,000

ROI=?

ROI is the return on investment that gives the loss or gain in any investment and is calculated by the following formula:

ROI= (interest/investment) x 100

Finding interest:

interests= received sum- investment

              = 20,000-12,000

              =8000

Putting values in ROI formula we get:

ROI= (8000/12000) x 100

      = 66.67%

Hence the return on investment is 66.67%!

3 0
3 years ago
Please help check my answers I beg you!!!!!!!
Alchen [17]
1. the answes for the graph ones is the 3rd and 4th graph because both graphs are 1 line and a line is infinite

2. (-3,5) x=(-) y=(+)that means that it must to the left and up part of graph, therefor it is the 4th graph

1. it would be the 4th graph that you didn't show because none of the included answers are correct
7 0
3 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                         T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean percentage of organic matter = 2.481%

             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

                                     =  -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 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 <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0
3 years ago
Solve for x.3-sqrt(x)=0 What is the root? If there is none, choose none. x = 3 x = 9 none
Anon25 [30]

Answer:

x = 9

Step-by-step explanation:

You can solve this problem by isolating the variable.

3-√x = 0  

√x = 3 (add √x to both sides of the equation)

x = 9 (square both sides of the equation)

8 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST<br> Answer all parts of the question please!
Zanzabum

Answer:

It's in the air for about 2 seconds (A), maximum height was about 17 feet, and yes you win the price since 17> 15.

Step-by-step explanation:

7 0
2 years ago
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