Question

p-value also

A recent broadcast of a television show had a 15 share, meaning that among 6000 monitored households with TV sets in use, 15% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in use, less than 25% were tuned into the program. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: p = 0.75 H:p<0.75 0 C. Hạ: p = 0.75 Hi:p>0.75 O E. Ho: p = 0.25 Hi:p>0.25 O B. Ho: p = 0.75 H: P0.75 0 D. Ho: p = 0.25 Hy: P0.25 F. Họ: p = 0.25 H: P<0.25 The test statistic is z= und to two decimal places as needed.)

Answer 1

Solution :

Given that,

n = 6000

$url=http://latex.codecogs.com/gif.latex?%5Chat%20p&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=W8lDZu9THY8EdwqflwtFkA--~C$ = 0.15

Hypothesis:

H0 : P $=$ 0.25   

Ha : P 0.25

Test Statistic:

Z =  $url=http://latex.codecogs.com/gif.latex?%5Chat%20p&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=W8lDZu9THY8EdwqflwtFkA--~C$ - $\hat{}$ $\hat{}$P / ( $url=http://latex.codecogs.com/gif.latex?%5Csqrt&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=wJ2iHy0.CBXNpMXEBQTPPg--~C$ ((P * (1 - P)) / n)

  Z = 0.15 - 0.25 / ( $url=http://latex.codecogs.com/gif.latex?%5Csqrt&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=wJ2iHy0.CBXNpMXEBQTPPg--~C$ (( 0.25 * 0.75) / 6000)

Z = -17.89

Z = -17.89

P-value = 0.0000

Decision Region : If P-value $\leq$    $url=http://latex.codecogs.com/gif.latex?%5Calpha&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=aMjjgR9sgJwjyI3pJyZRDg--~C$ = 0.01 then reject Null hypothesis(H0)

P-value = 0.0000 $url=http://latex.codecogs.com/gif.latex?%5Calpha&t=1587225593&ymreqid=518bbb84-0bf4-d976-1c05-36000301c400&sig=aMjjgR9sgJwjyI3pJyZRDg--~C$ = 0.01, reject H0

There is sufficient evidence to support the claim.

An advertiser that among the households with TV sets in use , less than 25% were tuned into the proggram.