contestada

An admissions officer for a law school has determined that historically applicants have undergraduate grade point averages that are normally distributed with standard deviation 0.3. From a random sample of 25 applications from the current year, the sample mean grade point average is 3.2. (a)Find a 99% confidence interval for the population mean. (b)Find a 95% confidence interval for the population mean.(c)Find a 90% confidence interval for the population mean.(d)Based on these sample results, a statistician computes for the population mean a confidence interval extending from 3.15 to 3.25. Find the confidence level associated with this interval.

Respuesta :

zainsubhani

Answer:

Part A:

interval=(3.0452,3.3548,)

Part B:

Interval=(3.0824,3.3176)

Part C:

Interval=(3.100763.29924,)

Part D:

Z=0.833

CI=91%

Step-by-step explanation:

CI                                                          Z

90%                                                  1.645

95%                                                  1.96

99%                                                  2.58

The formula we are going to use is:

Interval=X±[tex]\frac{Z*S}{\sqrt{n} }[/tex]

Where

X is the mean value

S is the standard deviation

n is the sample size

Z is the distribution

Part A:

Interval=3.2±[tex]\frac{2.58*0.3}{\sqrt{25} }[/tex]

Interval=3.2±0.1548

interval=(3.0452,3.3548)

Part B:

Interval=3.2±[tex]\frac{1.96*0.3}{\sqrt{25} }[/tex]

Interval=3.2±0.1116

Interval=(3.0824,3.3176)

Part C:

Interval=3.2±[tex]\frac{1.654*0.3}{\sqrt{25} }[/tex]

Interval=3.2±0.09924

Interval=(3.10076,3.29924)

Part D:

3.2-3.15=3.25-3.2=0.05

0.05=[tex]\frac{Z*S}{\sqrt{n} }[/tex]

0.05=[tex]\frac{Z*0.3}{\sqrt{25} }[/tex]

Z=0.833

CI=91%

Otras preguntas