Question

The Nielsen Company is a global information and media company and one of the leading suppliers of media information. In their state-of-the-media report, they announced that U.S. cell phone subscribers average 5.4 hours per month watching videos on their phones. We decide to construct a 95% confidence interval for the average time (hours per month) spent watching videos on cell phones among U.S. college students. We draw the following SRS of size 8 from this population.

11.8    2.7    3.0    6.1    4.8    9.8    11.2    7.7

We want to test whether the average time that U.S. college students spend watching videos on their phones differs from the reported overall U.S. average at the 0.05 significance level. Specifically, we want to test the following.

H0: μ	=
Ha: μ	≠

From this sample, we have the following. (Round your answers to two decimal places.)

n	=
x	=
s	=

The t statistic is calculated below. (Round your answers to two decimal places.)

t

=

x − μ₀

s /

n

=

− 5.4

  

=

This means that the sample mean

x =

is slightly less than 1.5 standard deviations away from the null hypothesized value

μ =  .

Because the degrees of freedom are

n − 1 =  ,

this t statistic has the

t

distribution. The figure below shows that the P-value is

2P

T ≥

,

where T has the t(7) distribution. From Table D we see that

P(T ≥ 1.119) =

and

P(T ≥ 1.415) =  .

P-value = 0.2106

t

t

Therefore, we conclude that the P-value is between

2 ✕ 0.10 =

and

2 ✕ 0.15 =  .

Software gives the exact value as

P-value = 0.2106.

These data are compatible with a mean of  hours per month. Under H₀, a difference this large or larger would occur about one time in five simply due to chance. There is  ---Select--- is is not enough evidence to reject the null hypothesis at the 0.05 level.

Answer 1