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Ugo [173]
3 years ago
11

Jim earns 1,600 per month after taxes. he is working on his budget and has the first threecategories finished. Housing $576 Food

$272 Transportation $320
Mathematics
1 answer:
k0ka [10]3 years ago
4 0
Housing + Food + Transportation= 1,168
1,600 - 1,168= 432 left.
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Raul is making a scale model of an airplane that has a wingspan of 44 feet. If Rauls scale model is 1/16 the size of the actual
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You would divide 44 by 16 so the wing span is 2.75 ft.
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What percentage of 70 is 49?<br> Your answer
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the answer to your question is 70%. 49 is 70% of 70

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2 years ago
A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01
slavikrds [6]

Answer:

a) z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649  

b) For this case we need to find a critical value that accumulates \alpha/2 of the area on each tail, we know that \alpha=0.01, so then \alpha/2 =0.005, using the normal standard table or excel we see that:

z_{crit}= \pm 2.58

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

\hat p=\frac{174}{594}=0.293 estimated proportion of yellow peas

p_o=0.23 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 of yellow peas is 0.23:  

Null hypothesis:p=0.23  

Alternative hypothesis:p \neq 0.23  

When we conduct a proportion test we need to use the z statisitc, 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{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649  

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=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>3.649)=0.00026  

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.

b) Critical value

For this case we need to find a critical value that accumulates \alpha/2 of the area on each tail, we know that \alpha=0.01, so then \alpha/2 =0.005, using the normal standard table or excel we see that:

z_{crit}= \pm 2.58

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

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3 years ago
During a catered lunch an average of 4 cups of tea are poured per minute. This lunch will last 2 hours. How many gallons of tea
zzz [600]
4 cups are poured every minute...there are 60 minutes in 1 hr...so for 2 hrs, there is 120 minutes...at 4 per minute = 120 * 4 = 480 cups in 2 hrs

16 cups = 1 gal
480 cups = 480/16 = 30 gallons <===
4 0
3 years ago
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