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Anit [1.1K]
3 years ago
13

In CAT entrance examination paper there are 3 sections, each containing 5 questions. A candidate has to solve 5,

Mathematics
1 answer:
mamaluj [8]3 years ago
4 0
The best and most correct answer providedfrom your question about the entrance examination paper is the second option which is 2,250. The problem can be solved by:

3-1-1 : 5c3*5c1*5c1 *3!/2! = 750 
<span>2-2-1 : 5c2*5c2*5c1 *3!/2! = 1500
</span>
Adding the two answers:

750 + 1500 = 2250

I hope it has come to your help.


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Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that th
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The null hypothesis is H_0: \mu = 14

The alternate hypothesis is H_a: \mu < 14

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

H_0: \mu = 14

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

H_a: \mu < 14

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that \mu = 14

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that n = 40

Test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}

t = -1.95

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

5 0
2 years ago
NEED HELP ASAP PLEASE
Keith_Richards [23]
It’s the second answer
<1=58 <2=122 <3=58
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3 years ago
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