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zavuch27 [327]
2 years ago
13

The foreman in the welding department wanted to know what value of allowance to use for a particular section of the shop. A work

sampling study was authorized. Only two activity categories were considered: (1) welding and other productive work, (2) personal time, rest breaks, and delays. Over a four week period (40 hours/week), 125 observations were made at random times. Each observation captured the category of activity of each of eight welders in the shop section of interest. Results indicated that category 2 constituted 33% of the total observations. (a) Define the limits of a 96% confidence interval (SD
Mathematics
1 answer:
frez [133]2 years ago
3 0

Answer:

The answer is "0.3605".

Step-by-step explanation:

Through 8 welds, every welder have 125 measurements, that overall amount of measurements.  

N= 125(8) = 1000  

The 96 percent level of trust

Z_{\frac{\alpha}{2}} =2.05\\\\\hat{\sigma}_{p} = \sqrt{\frac{0.33 (0.67)}{1000}}\\\\

    =\sqrt{ \frac{0.2211}{1000}}\\\\=\sqrt{0.0002211}\\\\=0.01487

\hat{P_2} - Z_{\frac{\alpha}{2}} \hat{\sigma}_{p} = 0.33 -2.05 (0.01487)\\\\

                  = 0.33-0.0305\\\\=0.2995

\hat{P_2} + Z_{\frac{\alpha}{2}} \hat{\sigma}_{p} = 0.33 +2.05 (0.01487)

                  = 0.33+0.0305\\\\=0.3605

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Check out the diagram below. Figure 1 is what she already has. Figure 2 is what happens after completing the next step. The red and blue arcs intersect to help form the endpoints of the perpendicular bisector. I used GeoGebra to make the diagrams.

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The class president is organizing a class trip to a nearby amusement park for 314 students. The regular price is $35 per ticket.
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287 students were eligible for the discount :)

Step-by-step explanation:

7 0
2 years ago
According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother
k0ka [10]

Answer:

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

p_v =P(z>7.124)=5.24x10^{-13}  

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, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.  

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

\hat p=\frac{331}{502}=0.659 estimated proportion of people who would not be bothered if the NAtional security agency

p_o=0.5 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 the majority of 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

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{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

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>7.124)=5.24x10^{-13}  

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, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.  

4 0
3 years ago
A culture started with 6,000 bacteria. After 5 hours, it grew to 7,800 bacteria. Predict how many bacteria will be present after
AleksandrR [38]

Answer:

After 5 hours, the amount of bacteria increased: 7800-6000=1800

After 1 hours, the amount of bacteria increased: 1800/5=360

After 18 hours,  the amount of bacteria increased: 360*18 = 6480

=> Total amount of bacteria after 13 hours: 6000+6480 =12480

4 0
3 years ago
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