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mojhsa [17]
3 years ago
15

It is believed that 11% of all Americans are left-handed. In a random sample of 100 students from a particular college with 1012

3 students, 14 were left-handed. Find a 98% confidence interval for the percentage of all students at this particular college who are left-handed.
Mathematics
1 answer:
tia_tia [17]3 years ago
6 0

Answer:

Step-by-step explanation:

We can calculate this confidence interval using the population proportion calculation. To do this we must find p' and q'

Where p' = 14/100= 0.14 (no of left handed sample promotion)

q' = 1-p' = 1-0.14= 0.86

Since the requested confidence level is CL = 0.98, then α = 1 – CL = 1 – 0.98 = 0.02/2= 0.01, z (0.01) = 2.326

Using p' - z alpha √(p'q'/n) for the lower interval - 0.14-2.326√(0.14*0.86/100)

= -2.186√0.00325

= -2.186*0.057

= 12.46%

Using p' + z alpha √(p'q'/n)

0.14+2.326√(0.14*0.86/100)

= 0.466*0.057

= 26.5%

Thus we estimate with 98% confidence that between 12% and 27% of all Americans are left handed.

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Ephraim took a friend for a birthday dinner. The total bill for dinner was $30.29 (including tax and a tip). If Ephraim paid a 2
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Answer:

x= 30.29 (1-0.223) = 23.54

Ephraim paid a 22.3% tip, what was his bill before adding the tip?

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Step-by-step explanation:

For this case we assume the following notation:

T= 30.29 represent the total paid for the dinner

x represent the amount before adding the tip

f =0.223 represent the fraction of the total amount destined to the tip

Based on this we can create the following relationship between T and x:

T = f*T + (1-f)T = f*T + x

And we can solve for x from the previous equation like this:

x = T - f*T

Taking T common factor we got:

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