In: Statistics and Probability
A recent broadcast of a television show had a 10 share, meaning that among 6000 monitored households with TV sets in use, 10% 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 15% were tuned into the program. Use the P-value method. Use the normal distribution as anapproximation of the binomial distribution
A) Identify the null hypothesis, and alternative hypothesis
B) test statistic z= (round to nearest 2 decimal places)
C) P-value (4 decimal places)
D) make a conclusion about the null hypothesis (reject or fail to reject, is or is not sufficeint evedince to support the claim of 15%)
Given,
6000 monitored households with TV sets in use, 10% of them were tuned to this program
claim of an advertiser that among the households with TV sets in use, less than 15% were tuned into the program
A) Hypothesied proportion : = 15/100 = 0.15
Null hypothesis : Ho :
Alternate hypothesis : Ha : ( Left Tailed test)
B)
Given | |
Hypothesied Proportion : | 0.15 |
Sample Size | 6000 |
Sample proportion = 10/100 = 0.1
Level of significance = 0.01
Z = - 10.8465
p-value
C)
D)
As P-Value i.e. is less than Level of significance i.e (P-value:0 < 0.01:Level of significance); Reject Null Hypothesis
is sufficient evedince to support the claim of 15%
Using Excel Function to find the P-value : NORM.S.DIST functionNORM.S.DIST function
Returns the standard normal distribution (has a mean of zero and
a standard deviation of one).
Use this function in place of a table of standard normal curve
areas.Syntax – Standard Normal
DistributionNORM.S.DIST(z,cumulative)
The NORM.S.DIST function syntax has the following arguments:
## Z Required. The value for which you want the distribution.
## Cumulative Required. Cumulative is a logical value that
determines the form of the function. If cumulative is TRUE,
NORMS.DIST returns the cumulative distribution function; if FALSE,
it returns the probability mass function.