Denise spent $3.45 on snacks every day for 11 days. What is the amount of money Denise spent on these snacks?
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Answer:
A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].
Step-by-step explanation:
We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.
For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.
Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;
P.Q. = ~
where, = sample mean speed for the 25-mil film = 1.15
= sample mean speed for the 20-mil film = 1.06
= sample standard deviation for the 25-mil film = 0.11
= sample standard deviation for the 20-mil film = 0.09
= sample of 25-mil film = 8
= sample of 20-mil film = 8
= population mean speed for the 25-mil film
= population mean speed for the 20-mil film
Also, =
= 0.1005
<em>Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.</em>
<u>So, 98% confidence interval for the difference in population means, (</u><u>) is;</u>
P(-2.624 < < 2.624) = 0.98 {As the critical value of t at 14 degrees of
freedom are -2.624 & 2.624 with P = 1%}
P(-2.624 < < 2.624) = 0.98
P( <
< ) = 0.98
P( < (
) <
) = 0.98
<u>98% confidence interval for</u> () = [
,
]
= [ ,
]
= [-0.042, 0.222]
Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].
Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.