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Andre45 [30]
3 years ago
13

Watching tv: in 2012, the general social survey asked a sample of 1310 people how much time they spend\t watching tv each day. T

he 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 =
Mathematics
1 answer:
gayaneshka [121]3 years ago
6 0

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  

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  

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.  

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<img src="https://tex.z-dn.net/?f=%2826%20%5Cdiv%20100%29%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5Ctimes%2010" id
taurus [48]

Answer:

\boxed{\bf \:  \cfrac{13}{5}}

<u>Or in Decimal:</u>

\boxed{\bf \: 2.6}

Step-by-step explanation:

<u>Given expression :-</u>

\sf \: ( 26 \div 100) \times 10

<u>Solution :-</u>

\sf  = (26 \div 100 )\times 10

This arithmetic expression may be rewritten as ;

\sf  =  \cfrac{26}{100}  \times 10

Step 1 : <u>Cancel the zero of 10 and one zero of 100</u> :-

\sf  =  \cfrac{26}{10 \cancel0}  \times 1 \cancel0

<em>Results to;</em>

\sf  =  \:  \cfrac{26}{10}  \times 1

\sf  =  \:  \cfrac{26}{10}

Step 2: <u>Cancel 26 and 10</u><u> </u><u>by 2</u> :-

\sf  =  \cfrac{ \cancel{26}}{ \cancel{10}}

<em>Results to;</em>

\sf = \cfrac{ \cancel{26} {}^{13} }{ \cancel{10} {}^{5} }

\sf  =  \cfrac{13}{5}

<em>It can also be in Decimal.</em>

That is;

\sf = 2.6

Hence, the answer of the expression would be 13/5 or 2.6 .

\rule{225pt}{2pt}

I hope this helps!

Let me know if you have any questions.

I am joyous to help!

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