a. It is believed that today\'s teenagers use instant messaging on the Internet

Question

a. It is believed that today's teenagers use instant messaging on the Internet an average of 2 hours per day with a known population

standard deviation of 1/2 hour.A group of teachers thought that the average time spent was different than 2 hours per day. They

randomly surveyed 100 teenagers and calculated an average amount of time of 1.75 hours.Conduct a hypothesis test to determine if

the average amount of time teenagers spend using instant messaging is less than 2 hours per day.Use ? =1%.

b. Find a 99% confidence interval for the population mean time teens spend on the internet per day.

c. What is the error of this confidence interval you found in part b?

d. Does the confidence interval back up your response to the hypothesis test in part a? How do you know?

Explanation / Answer

a) Here we have to test the hypothesis that,

H0 : mu = 2 hours Vs H1 : mu < 2 hours

where mu is population mean amount of time

Assume alpha = level of significance = 1% = 0.01

Given that,

Sample mean (Xbar) = 1.75

population standard deviation (sigma) = 1/2 hours = 0.5 hours

Sample size (n) = 100

Here we use one sample z-test because sample size is > 30 and population standard deviation is known.

The test statistic is,

Z = (Xbar - mu) / (sigma / sqrt(n))

= (1.75 - 2) / (0.5/sqrt(100)) = -5

Now we have to find P-value for taking decision.

P-value we can find in excel.

syntax :

=NORMSDIST(z)

where z is z-score

P-value = 2.86652E-07 = 0.0000

P-value < alpha

Reject H0 at 1% level of significance.

Conclusion : There is sufficient evidence to say that the average amount of time teenagers spend using instant messaging is less than 2 hours per day.

b. Find a 99% confidence interval for the population mean time teens spend on the internet per day.

99% confidence interval for mu is,

Xbar - E < mu < Xbar + E

where E is margin of error.

E = (Zc * sd) / sqrt(n)

where Zc is critical value for normal distribution.

For 99% confidence Zc = 2.58

E = (2.58 * 0.5) / sqrt(100) = 0.13

Lower limit = Xbar - E = 1.75 - 0.13 = 1.62

Upper limit = Xbar + E = 1.75 + 0.13 = 1.88

99% confidence interval for the population mean time teens spend on the internet per day is (1.62, 1.88).

c. What is the error of this confidence interval you found in part b?

Margin of error = 0.13

d. Does the confidence interval back up your response to the hypothesis test in part a? How do you know?

We can see that Popultion mean = 2 hour is not lie in the confidence interval therefore we reject H0.

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