The board of directors of Saner Automatic Door Company consists of 12 members, 3 of whom are women. A new policy and procedures
1 answer:
Answer:
(a) The probability that all members of the committee are men is 0.3818.
(b) The probability that at least 1 member of the committee is a woman is 0.6182.
Step-by-step explanation:
The number of men in the board of directors is 9.
The number of women in the board of directors is 3.
A committee of three is randomly selected from the board.
(a)
Compute the probability that all members of the committee are men as follows:
P (Selecting 3 men) = n (Selecting 3 men) ÷ n (Selecting 3 members)
Thus, the probability that all members of the committee are men is 0.3818.
(b)
Compute the probability that at least 1 member of the committee is a woman as follows:
P (At least 1 woman) = 1 - P (All 3 men)
Thus, the probability that at least 1 member of the committee is a woman is 0.6182.
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Answer:
a) Number = 60 *0.35=21
b) Since is a left tailed test the p value would be:
c) If we compare the p value obtained and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.
We can use as smallest significance level 0.044 and we got the same conclusion.
Step-by-step explanation:
Data given and notation
n=60 represent the random sample taken
X represent the business owners plan to provide a holiday gift to their employees
estimated proportion of business owners plan to provide a holiday gift to their employees
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Part a
On this case w ejust need to multiply the value of th sample size by the proportion given like this:
Number = 60 *0.35=21
2) Part b
We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:
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
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about 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 provided . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
Part c
If we compare the p value obtained and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.
We can use as smallest significance level 0.044 and we got the same conclusion.