Question

13

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.6 hours.

(a)

The sample standard deviation was not reported, but suppose that
it was 5 hours. Carry out a hypothesis test with a significance
level of 0.05 to decide if there is convincing evidence that the
mean time spent using the Internet by Canadians is greater than
12.4 hours. (Use a statistical computer package to calculate the
*P*-value. Round your test statistic to two decimal places
and your *P*-value to three decimal places.)

*t*=_____

*P*-value= _____

State the conclusion in the problem context.

Do not reject *H*_{0}. We have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

Reject *H*_{0}. We have convincing evidence that
the mean weekly time spent using the Internet by Canadians is
greater than 12.4 hours.

Do not reject *H*_{0}. We do not have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

Reject *H*_{0}. We do not have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

(b)

Now suppose that the sample standard deviation was 2 hours.
Carry out a hypothesis test with a significance level of 0.05 to
decide if there is convincing evidence that the mean time spent
using the Internet by Canadians is greater than 12.4 hours. (Use a
statistical computer package to calculate the *P*-value.
Round your test statistic to two decimal places and your
*P*-value to three decimal places.)

*t*=____

*P*-value=_____

State the conclusion in the problem context.

Reject *H*_{0}. We have convincing evidence that
the mean weekly time spent using the Internet by Canadians is
greater than 12.4 hours.

Do not reject *H*_{0}. We do not have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

Reject *H*_{0}. We do not have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

Do not reject *H*_{0}. We have convincing
evidence that the mean weekly time spent using the Internet by
Canadians is greater than 12.4 hours.

(c)

Explain why the hypothesis tests resulted in different conclusions for part (a) and part (b).

The smaller standard deviation means that you can expect more
variability in measurements and greater deviations from the mean.
This explains why *H*_{0} is rejected when the
sample standard deviation is 2, but not when the sample standard
deviation is 5.

The larger standard deviation means that you can expect more
variability in measurements and greater deviations from the mean.
This explains why *H*_{0} is rejected when the
sample standard deviation is 2, but not when the sample standard
deviation is 5.

The larger standard deviation means that you can expect less
variability in measurements and smaller deviations from the mean.
This explains why *H*_{0} is rejected when the
sample standard deviation is 5, but not when the sample standard
deviation is 2.

The smaller standard deviation means that you can expect less
variability in measurements and smaller deviations from the mean.
This explains why *H*_{0} is rejected when the
sample standard deviation is 5, but not when the sample standard
deviation is 2.

Answer #1

Here in this scenario our claim is that the the mean time spent using the Internet by Canadians is greater than 12.4 hours.

To test this claim we have to use t distribution for one sample t test because here the population standard deviations is unknown. According to central limit theorem we can use z distribution instead of t but both gives us similarly results.

a)

Here we assumed that the sample Standerd deviation is 5 hours.

b) In this part we assumed that the sample Standerd deviation is 2 hours. Considering both cases we performed the one sample t test for both cases at 0.05 level of significance as below,

The t critical value is calculated using t table or using Excel.

Part a)

*t*=1.265

*P*-value= 0.1031

The p value is calculated using Excel.

conclusion in the problem context.

Since p value is greater than alpha level of significance 0.05. so,

We do not reject *H*_{0}. We do not have
convincing evidence that the mean weekly time spent using the
Internet by Canadians is greater than 12.4 hours.

At 0.05 level of significance our result is not significant.

Part b)

Here we assumed that the sample Standerd deviation is 2 hours then the ,

*t*= 3.16

*P*-value= 0.001.

the conclusion in the problem context.

Since p value is less than 0.05 we Reject
*H*_{0}. We have convincing evidence that the mean
weekly time spent using the Internet by Canadians is greater than
12.4 hours.

At 0.05 level of significance our result is significant.

c)

The larger standard deviation means that you can expect more
variability in measurements and greater deviations from the mean.
This explains why *H*_{0} is rejected when the
sample standard deviation is 2, but not when the sample standard
deviation is 5. This is appropriate explanation.

Other options are incorrect.

Thank you.

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