Question

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?

Answer 1

0.086%

0.533%

1.11%

2.22%

Answer 2

\[Error = \large \sigma_{\bar x}=\sigma/\sqrt n\]

Answer 3

which part goes where in the formula

Answer 4

i havent used that formula...not sure which part is which

Answer 5

n is usually the sample size

sigma is a known standard deviation ....

Answer 6

i think i might have read it wrong to start with ...

it seems to be asking how many standard deviation the mean is from the other mean

Answer 7

so n is 15 and sigma is 4.4?

Answer 8

\[\sqrt n~\frac{\mu_2-\mu_1}{\sigma}\]

Answer 9

let me read it once more .. these eyes are getting old on me

Answer 10

i am looking for margin or error.

Answer 11

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 26.4%.

What is the margin of error of the sample mean?


we can use the sample mean is a parameter for the population mean, and we alter the standard deviation by dividing the population sd by sqrt n

but there is usually a confidence interval that they speak of to formulate:\[CI=\hat p \pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}} \\~~~~~~~~~~~~~~~~------\\~~~~~~~~~~~~~~~~~error~margin\]

Answer 12

i dnt know how to plug in for all that

Answer 13

there might be a way to compare the .043 with \sqrt(.264(1-.264)/15)

but i cant recall any information that would suggest that

Answer 14

well, i dont know what you know so its kinda hard to approach a solution to this :)

Answer 15

http://www.dummies.com/how-to/content/how-to-calculate-the-margin-of-error-for-a-sample-.html

there is no z information to assess with

Answer 16

i can get an option by .43/sqrt(15) but i cant say with any certainty that it is correct. It just assumes a zscore of 1

Answer 17

my tutorial on it give me two formulas \[\sigma \div \sqrt{n}\] and sqrt of p(1-p)/n

Answer 18

the p parts deal with using the sample proportions as an approximation for the population proportions; but we are given the standard deviation of the population to begin with

Answer 19

\[\frac{.043}{\sqrt{15}}\approx \sqrt{\frac{.264(1-.264)}{15}}\]

Answer 20

yes but those are the only two formuls we are given

Answer 21

the o~ one gives us one of the options ...

Answer 22

.043/sqrt(15) would be my bet