Can anyone help. I really need it.
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Answer:
A 95% confidence interval for the true mean is [$3.39, $6.01].
Step-by-step explanation:
We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;
Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~
where, = sample mean income =
= $4.70
s = sample standard deviation = = $1.83
n = sample of parking meters = 10
= population mean
<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>
<u>So, 95% confidence interval for the population mean, </u><u> is ;</u>
P(-2.262 < < 2.262) = 0.95 {As the critical value of t at 9 degrees of
freedom are -2.262 & 2.262 with P = 2.5%}
P(-2.262 < < 2.262) = 0.95
P( <
<
) = 0.95
P( <
<
) = 0.95
<u>95% confidence interval for</u> = [
,
]
= [ ,
]
= [$3.39, $6.01]
Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].
The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.
Also, the margin of error =
= = <u>1.31</u>