Question

A random sample of 10 young adult men (20-30 years old) was sampled. Each person was asked how many minutes of sports they watched on television daily. The responses are listed below. Test the claim that the mean amount of sports watched on television by all young adult men is different from 50 minutes. Use a 1% significance level.

50 48 65 74 66 37 45 68 64 65

1. a) State the null (H0) and alternate (H1) hypotheses (indicate the claim).

2. b) Calculate the test statistic and P-value.

3. c) Make a decision to reject or fail to reject the null hypothesis.

4. d) Summarize the final conclusion in the context of the original claim.

Answer 1

Solution:

x	x2
50	2500
48	2304
65	4225
74	5476
66	4356
37	1369
45	2025
68	4624
64	4096
65	4225
∑x=582	∑x2=35200

Mean ˉx=∑xn

=50+48+65+74+66+37+45+68+64+65/10

=582/10

=58.2

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√35200-(582)2109

=√35200-33872.49

=√1327.69

=√147.5111

=12.1454

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :    = 50

H_a :     50

Test statistic = t

= ( - ) / S / n

= (58.2-50) /12.14 / 10

= 2.136

Test statistic = t = 2.136

P-value = 0.0614

= 0.01

P-value >

0.0614 > 0.01

Fail to reject the null hypothesis .

There is not sufficient evidence to claim that the population mean μ is different than 50, at the 0.01 significance level.