Question

Suppose data about the average number of hours people spend on social media websites at work is randomly collected for 5 days:

4, 4.5, 2.5, 2, 6

A.) If we assume this population is normally distributed, construct an 80% confidence interval for the population average amount of time people spend on social media websites at work.

B.) Suppose I want to know more about how workers spend their time throughout the day. I want to test the claim that people spend no more than 3 hours a day during their workday on Social Media. Test this claim (significance level =0.05).

(please show all work)

Answer 1

(A)

From the given data,the following statistics are calculated:

n = 5

$\bar{x}$ = 19/5 = 3.8

s =1.6047

$\alpha$ = 0.20

df = 5 - 1 = 4

From Table, critical values of t_C = $\pm$ 1.533

Confidence Interval:

T to x 8 -= 3.8 + Vn 1.533 x 1.6047 -= 3.8 – 1.1 = (2.7, 4.9)

So,

Answer is:

(2.7, 4.9)

(b)

H0: Null Hypothesis: $\mu$ $\leq$ 3 ( people spend no more than 3 hours a day during their workday on Social Media ) (Claim)

HA: Alternative Hypothesis:$\mu$ > 3 ( people spend more than 3 hours a day during their workday on Social Media )

$\bar{x}$ = 3.8

s = 1.6047

n = 5

$\alpha$ = 0.05

df = n - 1 = 5 - 1 = 4

One Tail - Right Side Test

From Table, critical value of t = 2.132

Test Statistic is given by:

t= T-Mo s/n 3.8 - 3 1.6047/5 = 1.115 1.6047/

Since calculated value of t = 1.115 is less than critical value of t = 2.132, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data support the claim that people spend no more than 3 hours a day during their workday on Social Media .