Is -93/6 equivalent to -93/6?

Plz answer............................

hjlf

Step-by-step explanation:

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3 0

3 years ago

LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last season. During the summer, h

aleksandr82 [10.1K]

Answer:

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$

$p_v =P(z>2.905)=0.0018$

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 free throws made is higher than 0.4

Step-by-step explanation:

Data given and notation

n=40 represent the random sample taken

X=25 represent the number of free throws made

$\hat p=\frac{25}{40}=0.625$ estimated proportion of free throws made

$p_o=0.4$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.4:

Null hypothesis: $p\leq 0.4$

Alternative hypothesis: $p > 0.4$

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:

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$

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:

$p_v =P(z>2.905)=0.0018$

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 free throws made is higher than 0.4

8 0

3 years ago

BRAINLIEST WILL BE GIVEN!! The budget for a school dance is $370.00. Decorations and refreshments cost $165.00. The band charges

soldi70 [24.7K]

Answer:

3 hours

Step-by-step explanation:

8 0

3 years ago

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