Suppose that a number of time teenagers spend playing computer games per week is

Question

Suppose that a number of time teenagers spend playing computer games per week is normally distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean computed as 6.5 hours.

a. Determine the 95% confidence interval estimate of the population mean.
b. Interpret the 95% confidence interval for this situation.
c. Determine and interpret the 99% confidence interval estimate of the population mean.
d. Determine and interpret the 90% confidence interval estimate of the population mean.
e. Determine the 95% confidence interval estimate of the population mean if the sample size is changed to 300.
f. Determine the 95% confidence interval estimate of the population mean if the sample size is changed to 36.
g. Determine the 95% confidence interval estimate of the population mean if the population standard deviation is changed to 2.
h. Determine the 95% confidence interval estimate of the population mean if the population standard deviation is changed to 1.2.
i. Determine the 95% confidence interval estimate of the population mean if the sample mean is changed to 5.0 hours.
j. Determine the 95% confidence interval estimate of the population mean if the sample mean is changed to 8.5 hours.

Explanation / Answer

Given,

Sample mean = 6.5

Sample standard deviation = 1.5

Sample size = 100

Question a)

95% confidence interval:

Confidence Interval:

Sample Mean (-/+) Margin of Error

Margin of Error = Critical value * (standard deviation / sqrt (sample size))

From Normal table we find the critical value for 5% level of significance as 1.96

6.5 (-/+) (1.96*(1.5/sqrt(100))

6.5 (-/+) 0.294

6.21 and 6.79

The 95% confidence interval is (6.21 and 6.79)

Question b)

Here we are 95% confident that the population mean of time teenagers spend playing computer games per week lies in between 6.21 hrs and 6.79 hrs.

Question c)

99% confidence interval:

Confidence Interval:

Sample Mean (-/+) Margin of Error

Margin of Error = Critical value * (standard deviation / sqrt (sample size))

From Normal table we find the critical value for 1% level of significance as 2.58

6.5 (-/+) (2.58*(1.5/sqrt(100))

6.5 (-/+) 0.387

6.11 and 6.89

The 99% confidence interval is (6.11 and 6.89).

Question d)

90% confidence interval:

Confidence Interval:

Sample Mean (-/+) Margin of Error

Margin of Error = Critical value * (standard deviation / sqrt (sample size))

From Normal table we find the critical value for 10% level of significance as 1.645

6.5 (-/+) (1.645*(1.5/sqrt(100))

6.5 (-/+) 0.247

6.25 and 6.75

The 90% confidence interval is (6.25 and 6.75).

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