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2 years ago

10

in a change jar, tom has nickels and dimes. the number of nickels is two times the number of dimes and the total amount of money

in the jar is $1.20. how many dimes are in the jar? a) 6 b)12 c)10 d)24

1 answer:

AleksAgata [21]2 years ago

8 0

N=number of nickles
d=number of dimes

n is 2 times d
n=2d
total money is 1.20
convert to cens
120 cetns

5n+10d=120
n=2d
5(2d)+10d=120
10d+10d=120
20d=120
divide both sides by 120
d=6
6 dimes is answer

answer is A, 6 dimes

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Answer:

i:

The appropriate null hypothesis is $H_0: p \geq 0.2$

The appropriate alternative hypothesis is $H_1: p < 0.2$

The p-value of the test is 0.1057 > 0.05, which means that there is not sufficient evidence that fewer than 20% of the museum visitors make use of the device, and so, it should not be withdrawn.

ii:

The p-value of the test is 0.1057

Step-by-step explanation:

Question i:

The device will be withdrawn if fewer than 20% of all of the museum’s visitors make use of it.

At the null hypothesis, we test if the proportion is of at least 20%, that is:

$H_0: p \geq 0.2$

At the alternative hypothesis, we test if the proportion is less than 20%, that is:

$H_1: p < 0.2$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

0.2 is tested at the null hypothesis:

This means that $\mu = 0.2, \sigma = \sqrt{0.2*0.8} = \sqrt{0.16} = 0.4$ .

The device will be withdrawn if fewer than 20% of all of the museum’s visitors make use of it. Of a random sample of 100 visitors, 15 chose to use the device.

This means that $n = 100, X = \frac{15}{100} = 0.15$

Test statistic:

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$z = \frac{0.15 - 0.20}{\frac{0.4}{\sqrt{100}}}$

$z = -1.25$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.15, which is the p-value of z = -1.25.

Looking at the z-table, z = -1.25 has a p-value of 0.1057.

The p-value of the test is 0.1057 > 0.05, which means that there is not sufficient evidence that fewer than 20% of the museum visitors make use of the device, and so, it should not be withdrawn.

Question ii:

The p-value of the test is 0.1057

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3 years ago

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Answer:

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b) $P(13.848$ Step-by-step explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

For this case we assume that the sample variance is given by $S^2$ and we select a random sample of size n from a normal population with a population variance $\sigma^2$ . And we define the following statistic:

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For this case we want this probability:

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$P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"

Part b

For this case we want this probability:

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If we multiply the inequality by 4 on all the terms we got:

$P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2$ And we can find this probability like this: $P(13.848$ And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"

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