The CEO of a local tech company wants to estimate the proportion of employees that are pleased with the cleanliness of the break
1 answer:
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Answer:
z= -0.968
We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .
Step-by-step explanation:
1) Data given and notation n
n=106 represent the random sample taken
X=48 represent the people who says that they would watch one of the television shows.
estimated proportion of people who says that they would watch one of the television shows.
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that 50% of people who says that they would watch one of the television shows.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level is not provided, but we can assume . The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
So based on the p value obtained and using the significance level assumed we have
so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they would watch one of the television shows not differs from 0.5 or 50% .