Question

Conduct a test at the

alphaαequals=0.100.10

level of significance by determining ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, and​ (c) the​ P-value. Assume the samples were obtained independently from a large population using simple random sampling.

Test whether

p 1 greater than p 2p1>p2.

The sample data are

x 1 equals 118x1=118​,

n 1 equals 256n1=256​,

x 2 equals 144x2=144​,

and

n 2 equals 319n2=319.

(a) When comparing two population​ proportions, the null hypothesis is a statement of​ "no difference." The alternative hypothesis is based on what is to be determined from the test. Recall that the null hypothesis is a statement of​ equality, and the alternative hypothesis is a statement of inequality. Choose the correct null and alternative hypotheses below. A. Upper H 0 : p 1 equals p 2 versus Upper H 1 : p 1 greater than p 2 Your answer is correct.B. Upper H 0 : p 1 equals 0 versus Upper H 1 : p 1 not equals 0 C. Upper H 0 : p 1 equals p 2 versus Upper H 1 : p 1 less than p 2 D. Upper H 0 : p 1 equals p 2 versus Upper H 1 : p 1 not equals p 2 ​(b) If the samples are independently obtained using simple random​ sampling, n 1 ModifyingAbove p with caret 1 left parenthesis 1 minus ModifyingAbove p with caret 1 right parenthesis greater than or equals 10 and n 2 ModifyingAbove p with caret 2 left parenthesis 1 minus ModifyingAbove p with caret 2 right parenthesis greater than or equals 10​, and the sample sizes are no more than​ 5% of the​ population, then the test statistic z 0 can be found. The problem assumes that the samples were obtained independently from a large population using simple random sampling. To make sure the formula can be​ used, evaluate n 1 ModifyingAbove p with caret 1 left parenthesis 1 minus ModifyingAbove p with caret 1 right parenthesis and n 2 ModifyingAbove p with caret 2 left parenthesis 1 minus ModifyingAbove p with caret 2 right parenthesis. Start by evaluating ModifyingAbove p with caret 1. Use the fact that ModifyingAbove p with caret Subscript i Baseline equals StartFraction x Subscript i Over n Subscript i EndFraction to evaluate ModifyingAbove p with caret 1. ModifyingBelow ModifyingAbove p with caret 1 With font size decreased by 2 nothing equals StartFraction x 1 Over n 1 EndFraction equals StartFraction 118 Over 256 EndFraction equals . 4609 ​(Round to four decimal places as​ needed.) Now evaluate ModifyingAbove p with caret 2. ModifyingBelow ModifyingAbove p with caret 2 With font size decreased by 2 nothing equals StartFraction x 2 Over n 2 EndFraction equals StartFraction 144 Over 319 EndFraction equals . 4514 ​(Round to four decimal places as​ needed.) Evaluate the condition for the first sample. Substitute and simplify. n 1 ModifyingAbove p with caret 1 left parenthesis 1 minus ModifyingAbove p with caret 1 right parenthesis equals 256 times 0.4609 left parenthesis 1 minus 0.4609 right parenthesis equals 63.61 ​(Round to two decimal places as​ needed.) Now substitute in the values for the second sample and simplify. n 2 ModifyingAbove p with caret 2 left parenthesis 1 minus ModifyingAbove p with caret 2 right parenthesis equals 319 times 0.4514 left parenthesis 1 minus 0.4514 right parenthesis equals 79.00 ​(Round to two decimal places as​ needed.) The formula for the test statistic is the following. z 0 equals StartFraction ModifyingAbove p with caret 1 minus ModifyingAbove p with caret 2 Over StartRoot ModifyingAbove p with caret left parenthesis 1 minus ModifyingAbove p with caret right parenthesis EndRoot times StartRoot StartFraction 1 Over n 1 EndFraction plus StartFraction 1 Over n 2 EndFraction EndRoot EndFraction where ModifyingAbove p with caret equalsStartFraction x 1 plus x 2 Over n 1 plus n 2 EndFraction Since both values are greater than​ 10, calculate the test statistic. While either the formula or technology can be used to calculate the test​ statistic, for this example use technology. z 0equals nothing ​(Round to two decimal places as​ needed.)

Answer 1

(a)

Hypotheses are:

(b)

Here we have

Pooled sample proportion is

Standard error of the test is:

So test statistics will be

Since test is right tailed so p-value is: 0.409

Excel function used for p-value: "=(1-NORMSDIST(0.49))"

Since p-value is greater than 0.10 so we fail to reject the null hypothesis.