Question

watching tv: in 2012, the general social survey asked a sample of 1310 people how much time they spend\t watching tv each day. The mean number of hours was 2.8 with a standard deviation of 2.6. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day is less the claim? Use the a=0.5 level of significance and the P-value method with the TI-84 Plus calculator.

(a) State the appropriate null and alternate hypotheses.

H0: Mu = 3
H1: Mu < 3

This hypothesis test is a _____ test.

(b) Compute the P-value. Round the answer to at least four decimal places

P-value =

Answer 1

Answer:

a) Null hypothesis:$\mu \geq 3$  

Alternative hypothesis:$\mu < 3$  

This hypothesis test is a left tailed test.

b) $t=\frac{2.8-3}{\frac{2.6}{\sqrt{1310}}}=-2.784$  

The p value for this case can be calculated with this probability:  

$p_v =P(z<-2.784)=0.0027$  

We can conduct the test with the Ti84 using the following steps:

STAT>TESTS>T-test>Stats

We input the value $\mu_o =3, \bar X= 2.8, s_x = 2.6, n=1310$ and for the alternative we select $< \mu_o$. Then press Calculate.

And we got the same results.  

Step-by-step explanation:

Information given

$\bar X=2.8$ represent the sample mean

$s=2.6$ represent the population standard deviation  

$n=1310$ sample size  

$\mu_o =3$ represent the value to test

$\alpha=0.5$ represent the significance level for the hypothesis test.  

t would represent the statistic

$p_v$ represent the p value for the test

Part a) System of hypothesis

We want to test if the true mean is less than 3, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 3$  

Alternative hypothesis:$\mu < 3$  

This hypothesis test is a left tailed test.

Part b

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

Replacing the info given we got:

$t=\frac{2.8-3}{\frac{2.6}{\sqrt{1310}}}=-2.784$  

The p value for this case can be calculated with this probability:  

$p_v =P(z<-2.784)=0.0027$  

We can conduct the test with the Ti84 using the following steps:

STAT>TESTS>T-test>Stats

We input the value $\mu_o =3, \bar X= 2.8, s_x = 2.6, n=1310$ and for the alternative we select $< \mu_o$. Then press Calculate.

And we got the same results.