Last year, 46% of business owners gave a holiday gift to their employees. a survey of business owners indicated that 35% plan to provide a holiday gift to their employees. suppose the survey results are based on a sample of 60 business owners. a. how many business owners in the survey planned to provide a holiday gift to their employees in 2009?

b. suppose the business owners in the sample did as they planned. compute the p-value for a hypothesis test that can be used to determine if the proportion of business owners providing holiday gifts had decreased from the 2008 level.

c. using a .05 level of significance, would you conclude that the proportion of business owners providing gifts decreased? what is the smallest level of significance for which you could draw such a conclusion?

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.  

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

step-by-step explanation: