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Westkost [7]
3 years ago
15

If there were 35,000 fatal traffic accidents, how many would be expected to involve drivers in the 55-64 age group?

Mathematics
1 answer:
olasank [31]3 years ago
5 0
You need more information to answer this question. Is there a table or percentage for that age group?
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18. A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be
avanturin [10]

Answer:

a. =50 ± 4.67

b. =50 ± 4.81  

decreasing the sample size increase the margin of error

c. =50 ± 3.51

decreasing the confidence level reduced the margin of error

d. No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

Step-by-step explanation:

given

The sample mean, x, = 50,

the sample standard deviation, s, = 8.

a. Construct a 98% confidence interval for m if the sample size, n, is 20

for a 98% confidence interval of an infinite population, thus for a sample of size N and mean x, drawn from an infinite population having a standard deviation of σ, the mean value of the population is estimated to by:

x± 2.33σ /√N

for a confidence level of 98%, Zc = 2.33 gotten from the table of confidence interval.

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√20)      and 50+ ( 2.33*8 /√20)

with 98% confidence in this prediction.

=50 ± 4.67

=45.33 or 54.67

(b) Construct a 98% confidence interval for m if the sample size, n, is 15.

using the same method as above

x± 2.33σ /√N

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√15)      and 50+ ( 2.33*8 /√15)

=50 ± 4.81

=45.19 or 54.81

decreasing the sample size increase the margin of error

(c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part

95% confidence interval =1.96 (gotten from table of confidence coefficient)

using x±1.96 σ /√N

This indicates that the mean value of the population lies between

50- (1.96 *8 /√20)      and 50+ (1.96 *8 /√20)

=50 ± 3.51

=46.49 or 53.51

c. decreasing the confidence level reduced the margin of error

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

7 0
3 years ago
How to write 100203 in word form??
sergij07 [2.7K]
One hundred thousand two hundred and three
8 0
4 years ago
Read 2 more answers
Can anyone please answer this question !!!
topjm [15]
Txuzitsidsujfistsjstssgj
7 0
3 years ago
suppose that you interview 1000 existing voters about who they voted for mayor. of the 1000 voters, 580 reported that they voted
pogonyaev

Answer:

420 is the answer of yhis lenthiest question u nedd to aubract them by 1000 and 580

Step-by-step explanation:

Just eubtract them

6 0
2 years ago
Write the equation of the line with slope of 2 and y-intercept (0, 15).
mr_godi [17]

Answer:

y = 2x + 15

Step-by-step explanation:

The slope-intercept form of the equation of a line is

y = mx + b

where m is the slope, and b is the y-coordinate of the y-intercept.

slope = m = 2

y-intercept = b = 15

y = 2x + 15

3 0
3 years ago
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