The company has 5 employees, who earn a total of $1,000 each working day. They are paid each Monday for their work in the five-d
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Answer:
A 98% confidence interval for the mean assembly time is [21.34, 26.49] .
Step-by-step explanation:
We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~
where, = sample average time = 23.92 minutes
s = sample standard deviation = 6.72 minutes
n = sample of times = 40
= population mean assembly time
<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>
<u>So, a 98% confidence interval for the population mean, </u><u> is;
</u>
P(-2.426 < < 2.426) = 0.98 {As the critical value of z at 1% level
of significance are -2.426 & 2.426}
P(-2.426 < < 2.426) = 0.98
P( <
<
) = 0.98
P( <
<
) = 0.98
<u>98% confidence interval for</u> = [
,
]
= [ ,
]
= [21.34, 26.49]
Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .