Question

1. It is believed that the population proportion of adults in the US who own dogs is 0.65. I surveyed people leaving the veterinarians office and found that 96 out of 150 owned a dog. Test this hypothesis at the .05 significance level. Assume a random sample.

2. Randomly surveyed 10 employees at work for their average on how many times they use the rest room per shift. the results were as follows 2,2,3,1,0,4,1,1,0,5

the mean for this is 1.9 times per shift.

test the hypothesis at a .5 significance level.

3.  In the company there are 845 employees who use laptops within the organization. In our office downtown in the city, there are 75 employees who use laptops, walking around in our downtown office there are 55 mac users and 20 dell users. Test this hypothesis with this random sample.

Answer 1

1. It is believed that the population proportion of adults in the US who own dogs is 0.65. I surveyed people leaving the veterinarians office and found that 96 out of 150 owned a dog. Test this hypothesis at the .05 significance level. Assume a random sample.

Answer)

Null hypothesis Ho : P = 0.65

Alternate hypothesis Ha : P is not equal to 0.65

N = 150

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 97.5

N*(1-p) = 52.5

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed p = 96/150

Claimed p = 0.65

N = 150

After substitution

Z = -0.26

From z table, p(z<-0.26) = 0.3974

But our test is two tailed

So, p-value is = 2* 0.3974 = 0.7948

As the obtained p-value is greater than 0.05 (given significance level)

We fail to reject the null hypothesis

And we have enough evidence to support the claim that the population proportion of adults in the US who own dogs is 0.65.