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vlabodo [156]
3 years ago
13

The ransin sports company has noted that the size of individual customer orders is normally distributed with a mean of $112 and

a variance of $9. if a soccer team of 16 players were to make the next batch of orders, what would be the standard error of the mean?
Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Standard error of the mean is computed by:

Standard error = SD/ sqrt N

Where:

N is the sample size

SD is the standard deviation

 

To get the standard deviation, you need to get the sqrt of the variance = sqrt 9 = 3

So plugging in our data:

 

Standard error = 3 / sqrt (16)

= 0.75

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Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba
chubhunter [2.5K]

Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

Step-by-step explanation:

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".  

p represent the real population proportion of interest  

\hat p =0.68 represent the estimated proportion

n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

4 0
4 years ago
Ive never did anything like this before, Can someone help?
FromTheMoon [43]

Answer:

A

Step-by-step explanation:

SA = πr² + πrs

r = 4.42

s = 2.61

SA = π(4.42)² + π(4.42)(2.61)

SA = 19.5364π + 11.5362π

SA = 61.34 + 36.22

SA = 97.56

The closest answer would be A.

Best of Luck!

7 0
3 years ago
Read 2 more answers
last week, Sara earned 84 dollars. she earned 14 dollars mowing lawns. she earned the rest of the money babysitting for 7 dollar
morpeh [17]
So she earned 84 total dollars
she earned 14 dollars mowing lawns
she earned (total-mowing lawns=babysitting) 84-14=70 dollar babysitting

7 dollars per hour so
70/7=10 so
h=10
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Jaclyn is selling cookies for her club. She sells 2 boxes for $5. What unit rate expresses the cost per box?
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Divide 5 by 2 representing each box by the price. Which equals to $2.50 each box.
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The correct option is d
7 0
3 years ago
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