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alexira [117]
3 years ago
14

Could someone please help answer this (homework help) will give brainliest, please no spam, thank you!

Mathematics
1 answer:
Nitella [24]3 years ago
4 0

Answer:

m= 4(25) + 2(7)

m= 100 + 14

m= 114

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The increasing cost of health care is an important issue today. Suppose that a random sample of 36 small companiesthat offer pai
garri49 [273]

Answer:

The 99% confidence interval is given by ($109.386;$138.613)

Step-by-step explanation:

1) Notation and definitions

n=36 represent the sample size

\bar X= 124 represent the sample mean  

s=32 represent the sample standard deviation

m represent the margin of error

Confidence =99% or 0.99

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".  

2) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. The degrees of freedom are given by:  

df=n-1=36-1=35  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.005,31)" for t_{\alpha/2}=-2.74  

"=T.INV(1-0.005,31)" for t_{1-\alpha/2}=2.74  

The critical value tc=\pm 2.74

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

m=t_c \frac{s}{\sqrt{n}}

m=2.74 \frac{32}{\sqrt{36}}=14.613

4) Calculate the confidence interval  

The interval for the mean is given by this formula:

\bar X \pm t_{c} \frac{s}{\sqrt{n}}

And calculating the limits we got:

124 - 2.74 \frac{32}{\sqrt{36}}=109.386

124 + 2.74 \frac{32}{\sqrt{36}}=138.613

The 99% confidence interval is given by ($109.386;$138.613)

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Answer:

the last answer... i hope this helps

Step-by-step explanation:

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