A recent study at UGA (March 5, 2008) suggests that students with cell phones may take more risks than students that do not have cell phones. In a random sample of 305 UGA female students, 128 responded that, if they had a cell phone, they would be willing to walk somewhere after dark that they would normally not go. Use the above survey results to test the claim that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go. H0: p = 0.50
HA: p < 0.50
(a) What is the value of the test statistic for this hypothesis test? (Use 2 decimal places in your answer).
(b) What is the correct conclusion for this hypothesis test at the 0.05 level of significance? (Multiple Choice: Is the answer 1, 2, 3, or 4?)
(1) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is sufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.
(2) Do not reject H0: p = 0.50; there is insufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.
(3) Do not reject H0: p = 0.50; there is sufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.
(4) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is insufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.

(a) -2.8

b) (1) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is sufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.

Step-by-step explanation:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

H0: p = 0.50

This means that:

In a random sample of 305 UGA female students, 128 responded that, if they had a cell phone, they would be willing to walk somewhere after dark that they would normally not go.

This means that

a) Value of the test statistic:

Pvalue:

We are testing the hypothesis that the proportion is less than 0.5, which means that the pvalue of the test is the pvalue of z = -2.81.

Looking at the z-table, z = -2.8 has a pvalue of 0.0026

(b) What is the correct conclusion for this hypothesis test at the 0.05 level of significance?

0.0026 < 0.05, which means that we reject the null hypothesis, that the proportion is 0.5, and accept the alternate hypothesis, that the proportion is less than 0.5, option (1).

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )