Mathematics, 21.06.2019 18:20 nehaljay1883

The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. if a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. what is the margin of error of the sample mean? a. 0.086% b. 0.533% c. 1.11% d. 2.22%

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Answer from: kaitlynhess

step-by-step explanation:

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Answer from: veronicacalyn

2.22%

Step-by-step explanation:

Answer from: NailahWalker4740

Step-by-step explanation:

Given that X the first-serve percentage of a tennis player in a match is normal

Std deviation =4.3%

Sample size n = 15

Std error of sample = std deviation/square root of n

= $\frac{4.3}{\sqrt{15} } =1.110%$

Margin of error =Z critical value x std error

If 95% confidence is considered then we have

Z critical value=1.96

Hence margin of error =1.96(1.11)%

=2.1756%

Answer from: creeconnell1

The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. The margin of error of the sample mean is 83.71%.

Answer from: nooooooooooooooo412

Given:
standard deviation = 4.3%
sample size = 15
mean = 26.4%

error = σ/√n

error = 4.3% / √15

error = 4.3% / 3.873

error = 0.0111

0.0111 x 100% = 1.11%

Answer from: raymond5799

It is now possible to find the margin of error, the reason being that the confidence level is not given. However the 'standard error' of the sample mean can be found as follows:
$Standard\ error=\frac{\sigma}{ \sqrt{n} }=\frac{0.043}{ \sqrt{15} }=0.011103$
Converted to a percentage, we get 1.11% which is the third choice of answer.

Answer from: nisslove

The formula of the margin of error of the sample mean.
MOE = standard deviation / square root (sample size).

MOE = 0.043 * 15 / square root (15)
MOE = 0.1665

So, the margin of error of the sample mean is 0.17

Answer from: vinp190p9zekn

We shall use the z-table to solve for the margin of error. The formula for the margin of error is
$E=z_{\alpha /2}\cdot \frac{\sigma }{\sqrt{n}}$

The value of z will be taken with the assumption that the significance level is 5%. So, α=0.025. Then $z_{\alpha \:/2}=1.96$ .

So, based from this, the margin of error is
$E=z_{\alpha /2}\cdot \frac{\sigma }{\sqrt{n}}=1.96\cdot \frac{0.043}{\sqrt{15}}=0.02$

The margin of error is about 2%.

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