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LenKa [72]
2 years ago
10

Help me solve this question!

Mathematics
1 answer:
MakcuM [25]2 years ago
5 0

Answer:

1st quartile is 136

The second quartile is also the median 142

The third quartile is 162

The interquartile range is the difference between the 3rd and first quartile or 26

Step-by-step explanation:

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Answer:

45 dogs

Step-by-step explanation:

so she has $550 and she did 20 cats and it costs $5 per cat and 5 x 20 is $100 and $550 - $100 = 450 and 450 divided by 10 ( because it's $10 per dog) is 45.

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Answer:

(1,-4) is in quadrant IV.

Step-by-step explanation:

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Coffee: A popular chain of cafes has been receiving online complaints about one store location. Regular customers complained tha
spin [16.1K]

Answer:

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\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 it's higher than 0.1 or 10%:  

Null hypothesis:p\leq 0.1  

Alternative hypothesis:p > 0.1  

When we conduct a proportion test we need to use the z statistic, 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{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

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 assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z>2.67)=0.0037  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .  

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3 years ago
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