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romanna [79]
3 years ago
14

Twelve different video games showing drug usedrug use were observed. The duration times of drug usedrug use were​ recorded, with

the times​ (seconds) listed below. Assume that these sample data are used with a 0.050.05 significance level in a test of the claim that the population mean is greater than 9090 sec. If we want to construct a confidence interval to be used for testing that​ claim, what confidence level should be used for a confidence​ interval? If the confidence interval is found to be 4.04.0 secless than
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
4 0

Answer:

a) The confidence level is 95%.

b) If the confidence interval, in seconds, is (4.0, 202.2) and the claim is that the population mean is greater than 90 seconds, we conclude that there is no enough evidence for that claim.

Step-by-step explanation:

a) If the hypothesis test is performed with a significance level of α=0.05, the confidence level that corresponds to this significance level is 95%:

CL=1-\alpha=1-0.05=0.95

If the confidence interval, in seconds, is (4.0, 202.2) and the claim is that the population mean is greater than 90 seconds, we conclude that there is no enough evidence for that claim.

The reason is that, with this confidence level, the confidence interval includes values that are below 90 seconds.

To have evidence to support that claim, the lower bound of the interval shouold have been greater than 90 seconds.

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In-s [12.5K]

Answer: it equals 20

Step-by-step explanation:

10 x 20 = 120, hoped this helped.

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ABCD FECG
MariettaO [177]

Answer:

x=12

Step-by-step explanation:

7 0
3 years ago
Plz help me with this
goldenfox [79]
Area of triangle=1/2(base×height)
360=1/2((4x-2)(7x+5))
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FOIL the binomial
720=28x^2+6x-10
Bring 720 to the other side
28x^2+6x-10-720=0
28x^2+6x-730=0
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3 years ago
If a 4-pound roast takes 150 minutes to cook, how long should a five-pound roast take
Ber [7]
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4 years ago
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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.

5 0
3 years ago
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