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3 years ago

4. The Crazy Candy Company ships out chocolates in boxes of 4 candies and boxes of 10

1 answer:

3 0

Using the least common multiple of 4 and 10, it is found that the smallest number of candies each order could have contained is 20 candies.

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There are boxes of 4 candies and of 10 candies.
Thus, the smallest possible number of equal candies on two orders in the<u> least common multiple of 4 and 10.</u>
To find this multiple, we factor both numbers by their prime factors, thus:

4 - 10|2

2 - 5|2

1 - 5|5

Thus, lcm(4,10) = 2*2*5 = 20.

The smallest possible number is 20 candies, one order with 20/4 = 5 boxes of 4 candies, and the other with 20/10 = 2 boxes of 10 candies.

A similar problem is given at brainly.com/question/24540608

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a) Null hypothesis: $p\leq 0.2$

Alternative hypothesis: $p > 0.2$

b) $z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25$

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Step-by-step explanation:

1) Data given and notation

n=400 represent the random sample taken

X=90 represent the number of questions with B as the correct answer

$\hat p=\frac{90}{400}=0.225$ estimated proportion of arrests that were not prosecuted

$p_o=0.2$ is the value that we want to test, since we assume that each question present 5 options and just one is correct, 1/5 =0.2 if all five options were equally likely

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:

Null hypothesis: $p\leq 0.2$

Alternative hypothesis: $p > 0.2$

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

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

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

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>1.25)=0.106$

If we compare the p value obtained and the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of B correct answers is not significantly higher than 0.2.