In a recent study of statistics students, a random sample of students were asked to provide the number of hours per week they spend studying for their statistics class. The results were used to compute confidence intervals for the population mean hours per week spent studying for statistics. The 95% confidence interval for the population mean hours per week that students spend studying for statistics was (6.88, 8.36).

In this confidence interval, what is the sample mean hours per week spent studying for statistics? Provide your answer as a number rounded to two decimal places.

2.If we want to create a 95% confidence interval for a population proportion with a margin of error (m) of 0.049, what sample size is necessary?

Use the simplified formula (n = 1m2 ) and be sure to round your final answer to the next largest whole number.

Provide your answer as a whole number (integer).

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!!!