Question

Question 1

As of January 2014, 58% of American adults have a Smartphone and on average, spend 34 hours per month using the mobile internet on Smartphone.

A researcher became interested to estimate the unknown average hours students in a given university spend in the mobile internet on Smartphone. The researcher takes a sample of 10 students in the university. The sample data is given below:

39

42

47

45

35

45

37

34

33

29

Suppose that the population standard deviation of students spending in the mobile internet on Smartphone is 7.89.

The point estimate of population mean μ  is

Group of answer choices

38.6

58

10

35

Question 2

As of January 2014, 58% of American adults have a Smartphone and on average, spend 34 hours per month using the mobile internet on Smartphone.

A researcher became interested to estimate the unknown average hours students in a given university spend in the mobile internet on Smartphone. The researcher takes a sample of 10 students in the university. The sample data is given below:

39

42

47

45

35

45

37

34

33

29

Suppose that the population standard deviation of students spending in the mobile internet on Smartphone is 7.89.

A 95% CI of the population mean μ  is

Group of answer choices

(34.49, 42.70)

34.51, 44.21

(31.71, 41.49)

(33.71, 43.49)

Question 3

A researcher wishes to find a 95% confidence interval for an unknown population mean μ using a sample of size 30. The population standard deviation is 8.88.

The margin of error for this confidence interval will be

Group of answer choices

3.2

1.6

2.7

0.95

Answer 1

Solution:

Question 1

Sample mean is an unbiased estimator of the population mean μ  

= (sum of observations)/n

= [39 + 42 + 47 + .........+ 29]/10

= 38.6

Answer : 38.6

Question 2

   = 7.89

c = 95% = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

The margin of error is given by

E =  /2 * ( / n )

= 1.96 * (7.89 / 10)

= 4.89

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(38.6 - 4.89)   <   <  (38.6 + 4.89)

33.71 <   < 43.49

Required 95% Confidence interval is

(33.71 , 43.49 )

Question 3

n = 300

   = 8.88

E =  /2 * ( / n )

= 1.96 * (8.88/ 30)

= 3.2

Answer : 3.2