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alina1380 [7]
3 years ago
9

Denise is a professional swimmer who trains, in part, by running. she would like to estimate the average number of miles she run

s in each week. for a random sample of 20 weeks, the mean is = 17.5 miles with standard deviation s = 3.8 miles. find a 99% confidence interval for the population mean number of miles denise runs. use a graphing calculator for this one and not the t chart from the book.
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0
Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval

The degrees of freedom is 
df = n-1 = 19

We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.

The 99% confidence interval is
xbar +/- t*(s/√n)

t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)

Answer: The 99% confidence interval is (15.07, 19.93)
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Step-by-step explanation:

We will take numbers to one side and variables to one side eliminate the "Square" with "Square Root" and solve for the variable x. The process is shown below:

3x^2-10=152\\3x^2=152+10\\3x^2=162\\x^2=\frac{162}{3}\\x^2=54\\x=\sqrt{54}

Now, we need to reduce the radical by using the rule shown below:

\sqrt{x*y}=\sqrt{x}  \sqrt{y}

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\sqrt{54}=\sqrt{6*9} =\sqrt{6}  \sqrt{9} =3\sqrt{6}

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Step-by-step explanation:

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Alik [6]

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<h3>What is a z-score?</h3>

The z-score is a numerical measurement used in statistics to refer to how much a given value differs from the standard deviation.

For the random variable x, the z-score is;

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therefore, 70.54% of the area under the curve lies to the right of x,

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