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Gwar [14]
3 years ago
12

What if the sample size used was some unspecified value larger than 30. Describe how this change in sample size would have affec

ted the following, if at all: The standard deviation of the distribution for the sample mean rent price for a one-bedroom apartment in this large city would ___________.
Mathematics
1 answer:
arlik [135]3 years ago
6 0

Answer:

Option D is correct.

Decreases.

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What is the greatest common factor of 60x4y7, 45x5y5, and 75x3y?
Burka [1]

Answer:

15x3y.

Step-by-step explanation:

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The distance DA between earth and the sun is 93,000,000 miles, the distance DE between earth and the moon is 240,000 miles, and
Rudiy27

Answer:

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Step-by-step explaidknation:

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A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% o
Semenov [28]
This can easily be answered. Justg use the next formula like this:
<span>0.4*0.05 = 0.02 
</span>So the probability would be of 2% of both shots
Hpe this is very useful
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3 years ago
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Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t
Snowcat [4.5K]

Answer:

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

\hat p=0.71 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.69 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 if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

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{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675  

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 is \alpha=0.05. 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>1.675)=0.047  

If we compare the p value obtained and using the significance level assumed \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 the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.  

5 0
3 years ago
THE QUESTION IS SOLVE FOR X​
Damm [24]

Answer:

34

Step-by-step explanation:

i honestly don't care i came on here to cheat but it's saying answer so here you go   :)

3 0
3 years ago
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