Respuesta :
Answer:
a) In this confidence interval, the sample mean hours per week spent studying for statistics = 7.62 hours per week
b) The sample size necessary for a 95% confidence interval for a population proportion with a margin of error (m) of 0.049 is approximately equal to 417.
Step-by-step explanation:
The confidence interval is given as
Confidence Interval = (Sample mean) ± (Margin of error)
The lower limit of the confidence interval
= (Sample mean) - (Margin of Error)
The upper limit of the confidence interval
= (Sample mean) + (Margin of Error)
Let the sample mean = x
Margin of error = y
The confidence interval from the question = (6.88, 8.36)
The lower limit of the confidence interval = 6.88 = x - y
The upper limit of the confidence interval = 8.36 = x + y
So, we have a simultaneous equation
x - y = 6.88
x + y = 8.36
2x = 15.24
x = 7.62
x + y = 8.36
y = 8.36 - 7.62 = 0.74
Sample Mean = 7.62
Margin of Error = 0.74
b) Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error)
Critical value for 95% confidence interval = 1.960 (Using the z-distribution)
Standard error = σₓ = √[p(1-p)/n]
where n = sample size = ?
p = sample proportion
But the sample proportion isn't given either.
But the the question instructs us to use the simplified formula
n = (1/m²)
m = Margin of error = 0.049
n = (1/0 049²) = 416.5 = 417 to the nearest whole number.
Hope this Helps!!!