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3 years ago

Can someome please help me?

Mathematics

2 answers:

natta225 [31]3 years ago

3 0

360 mm = 36 cm

Hope that helps

Licemer1 [7]3 years ago

3 0

1 cm= 10 mm
360÷10= 36cm Anoki bought

I hope this helped!!!:)

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2 years ago

A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n

mafiozo [28]

Answer:

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

$p_v =P(z>2.10)=0.018$

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Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

$\hat p=\frac{40}{120}=0.333$ estimated proportion of strawberries damaged

$p_o=0.25$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

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

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

$p_v =P(z>2.10)=0.018$

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

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2 years ago

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Answer:

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Step-by-step explanation:

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