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77julia77 [94]
2 years ago
9

A local health care company wants to estimate the mean weekly elder day-care cost. A sample of 10 facilities shows a mean of $25

0 per week, with a standard deviation of $25. What is the 90% confidence interval for the population mean?
Mathematics
1 answer:
laiz [17]2 years ago
8 0

Answer:

Step-by-step explanation:

Mean "X"=250

s = 25

n=10

Because Standard deviation is unknown and sample size is small, we must employ the T-distribution

df=n-1 = 10-1

df=9

For 90% confidence, t = 1.833

E=t*s/\sqrt{n\\ = 1.833*(25/\sqrt{10})

E=14.49

The 90% confidence interval is X±E=250±14.49 or 235.51, 264.49

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il63 [147K]
Hello there,

There is a total of 40 animals in that in 10 zebras, 10 lions, 10 elephants, and 10 monkeys.
So the probability of them both being lions is pretty rare. 2/40 so that would give you a 1/20 chance of both crackers being both lions.

I hope this helps!
7 0
2 years ago
Suppose that public opinion in a large city is 55% in favor of increasing taxes to support renewable energy and 45% against such
Harman [31]

Answer:

The approximate probability that more than 360 of these people will be against increasing taxes is P(Z> <u>0.6-0.45)</u>

                                              √0.45*0.55/600

The right answer is B.

Step-by-step explanation:

According to the given data we have the following:

sample size, h=600

probability against increase tax p=0.45

The probability that in a sample of 600 people, more that 360 people will be against increasing taxes.

We find that P(P>360/600)=P(P>0.6)

The sample proposition of p is approximately normally distributed mith mean p=0.45

standard deviation σ=√P(1-P)/n=√0.45(1-0.45)/600

If x≅N(u,σ∧∧-2), then z=(x-u)/σ≅N(0,1)

Now, P(P>0.6)=P(<u>P-P</u>   >     <u>0.6-0.45)</u>

                             σ          √0.45*0.55/600

=P(Z> <u>0.6-0.45)</u>

       √0.45*0.55/600

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2 years ago
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2 years ago
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