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gavmur [86]
3 years ago
11

A shelf for bowling balls can support 100 pounds. If a red bowling ball weighs 6 pounds and a blue bowling ball weighs 10 pounds

, can the shelf support 4 red bowling balls and 6 blue bowling balls? Write the total weight of the bowling balls.
Mathematics
2 answers:
sweet-ann [11.9K]3 years ago
8 0

Answer:

yes

84

Step-by-step explanation:

Llana [10]3 years ago
6 0

Answer:

yes, 84 pounds

Step-by-step explanation:

4 red balls would be 24 pounds

6 blue would be 60 pounds

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Answer:

z=\frac{0.802 -0.5}{\sqrt{\frac{0.5(1-0.5)}{121}}}=6.64  

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Step-by-step explanation:

Data given and notation

n=121 represent the random sample taken

X=97 represent the people who identify the color of the gummy bear

\hat p=\frac{97}{121}=0.802 estimated proportion of people who identify the color of the gummy bear

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Part a

For this case the parameter that we want to test is 0.5 for the proportion of the population of students will correctly identify the color

Part b: Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that there is no relationship between color and gummy bear flavor (population proportion different from 0.5).:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

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.

Part c: Calculate the statistic  

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

z=\frac{0.802 -0.5}{\sqrt{\frac{0.5(1-0.5)}{121}}}=6.64  

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=0.01. The next step would be calculate the p value for this test.  

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

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So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of students who will correctly identify the color is different than 0.5

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