Question

A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 10 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows.

14.2

15.8

22.8

23.1

20.7

37.6

13

13.9

20.5

20.0

Assume that the times spent on leisure activities by all adults are normally distributed. Using the 5% significance level, can you conclude that the average amount of time spent by American adults on leisure activities has changed? (Hint: First calculate the sample mean and the sample standard deviation for these data. Then make the test of hypothesis about μ.)

Round the sample standard deviation to three decimal places.

x¯=

s=

The claim is

.

Answer 1

necessary calculation table:-

hours(x)	x2
14.2	201.64
15.8	249.64
22.8	519.84
23.1	533.61
20.7	428.49
37.6	1413.76
13	169
13.9	193.21
20.5	420.25
20	400
sum=201.6	sum=4529.44

sample size (n) = 10

a).sample mean be:-

b).sample sd be:-

c).hypothesis:-

test statistic is:-

df = (n-1 )= (10-1) = 9

p value = 0.3669

[ in any blank cell of excel type =T.DIST.2T(0.95,9)]

decision:-

p value = 0.3669 >0.05

so we fail to reject the null hypothesis.

conclusion:-

there is not sufficient evidence  conclude that the average amount of time spent by American adults on leisure activities has changed at 0.05 level of significance.