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sammy [17]
3 years ago
10

In the week before and the week after a holiday there were 10,000 total deaths and 4968 of them occurred in the week before the

holiday.
a) construct a 90% confidence interval estimate of the proportion of deaths in the week before the holiday to the total death in the week before and the week after the holiday

b) based on the result does there appear to be any indication that people can temporarily postpone their death to survive the holiday
Mathematics
1 answer:
Norma-Jean [14]3 years ago
6 0
THIS MAY HELP YOU

CI = mean proportion ± (z*)sqrt[ (p)(1-p) / n]
where z* = 1.645 with 10%/2 = 0.05 area in each tail mean proportion p = 4946 / 10000 = 0.4946 now fill in the values from the problem ...
 CI = mean proportion ± (z*)sqrt[ (p)(1-p) / n]
 CI = 0.4946 ± (1.645)sqrt[ (0.4946)(1- 0.4946) / 10000]
 CI = (0.4864, 0.5028)

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Answer:

\% Change = \frac{9}{64} *100 = 14.0625\% \approx 14\%

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Step-by-step explanation:

For this case we assume that the info is given on the figure attached.

From this figure we have the following frequencies for each class

Class                  Frequency

___________________________

70.5-75.5                13

75.5-80.5                7

80.5-85.5                6

85.5-90.5                10

90.5-95.5                19

95.5-100.5               9

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Total                        64

So we have a total of 64 values and we want to find the percentage of students that scored higher than 95.5 so we can use the formula of relative change and we got:

\% Change = \frac{9}{64} *100 = 14.0625\% \approx 14\%

So then we have approximately 14% of the values higher than 95.5

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