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Vlada [557]
3 years ago
10

Dawn Dough and Dale earned different amounts $9.35,$8.52,and$8.25 dale and Doug earned about $9.00 Dawn earned exactly $1.10less

thsn Dale How much did each earn
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
8 0

We know that the amounts earned by Dawn, Doug and Dale are from the list of numbers: $9.35, $8.52 and $8.25

We also know that Dale and Doug earned close to $9.00

And that Dawn earned $1.10 less than Dale

Let the amount earned by Dale be x

⇒ Amount earned by Dawn is x - 1.1

If we notice the list of numbers, we see that $9.35 and $8.25 differ by $1.1

Hence, Dale earned $9.35 and Dawn earned $8.25

We are now left with $8.52, which should be the amount earned by Doug. This is correct, since we also know that Doug earned close to $9.

Hence, the amounts earned are:

Dale: $9.35

Doug: $8.52

Dawn: $8.25

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On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally dist
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Answer:

a) 8469-1.645\frac{100}{\sqrt{64}}=8448.4    

8469+1.645\frac{100}{\sqrt{64}}=8489.6    

So on this case the 90% confidence interval would be given by (8448.4;8489.6)    

b) 8469-2.054\frac{100}{\sqrt{64}}=8443.33    

8469+2.054\frac{100}{\sqrt{64}}=8494.68    

So on this case the 96% confidence interval would be given by (8443.33;8494.68)    

Step-by-step explanation:

Assuming this complete problem: On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with s = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of s.

(a) Assuming this to be the case, if a sample of 64 modified bars resulted in a sample average yield point of 8469 lb, compute a 90% CI for the true average yield point of the modified bar. (Round your answers to one decimal place.)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=8469 represent the sample mean

\mu population mean (variable of interest)

\sigma=100 represent the population standard deviation

n=64 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.645

Now we have everything in order to replace into formula (1):

8469-1.645\frac{100}{\sqrt{64}}=8448.4    

8469+1.645\frac{100}{\sqrt{64}}=8489.6    

So on this case the 90% confidence interval would be given by (8448.4;8489.6)    

(b) How would you modify the interval in part (a) to obtain a confidence level of 96%? (Round your answer to two decimal places.)

Since the Confidence is 0.96 or 96%, the value of \alpha=0.04 and \alpha/2 =0.02, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.02,0,1)".And we see that z_{\alpha/2}=2.054

Now we have everything in order to replace into formula (1):

8469-2.054\frac{100}{\sqrt{64}}=8443.33    

8469+2.054\frac{100}{\sqrt{64}}=8494.68    

So on this case the 96% confidence interval would be given by (8443.33;8494.68)    

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