Question

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?

Answer 1

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:  

$p_v =P(Z<-1.710)=0.0437$  

c) If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ 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

$\hat p=0.35$ estimated proportion of business owners plan to provide a holiday gift to their employees

$p_o=0.46$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ 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:$p\geq 0.46$  

Alternative hypothesis:$p < 0.46$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710$  

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 $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(Z<-1.710)=0.0437$  

Part c

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ 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.