The monthly rents for five apartments advertised in a newspaper were $670, $670, $790, $1550, and $810. Use the mean, median, mo
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Answer:
98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .
Step-by-step explanation:
We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;
Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95
Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60
Also, represent the population mean for Brand B and let
represent the population mean for Brand A.
Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;
P.Q. = ~
where, = Sample mean for Brand B data = 36.9
= Sample mean for Brand A data = 32.9
= Sample size for Brand B data = 13
= Sample size for Brand A data = 9
=
= 3.013
Here, and
are sample variance of Brand B and Brand A data respectively.
So, 98% confidence interval for the difference μX−μY is given by;
P(-2.528 < < 2.528) = 0.98
P(-2.528 < < 2.528) = 0.98
P(-2.528 * <
< 2.528 *
) = 0.98
P( (Xbar - Ybar) - 2.528 * <
< (Xbar - Ybar) + 2.528 *
) = 0.98
98% Confidence interval for μX−μY =
[ (Xbar - Ybar) - 2.528 * , (Xbar - Ybar) + 2.528 *
]
[ ,
]
[ 0.697 , 7.303 ]
Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .