Question

Conduct a test at the alphaequals0.05 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 2. The sample data are x 1 equals 122​, n 1 equals 244​, x 2 equals 137​, and n 2 equals 311.

Answer 1

There is not enough evidence to reject the null hypothesis.

Explanation:

(a)

The hypothesis can be defined as follows:

H₀: p₁ - p₂ ≤ 0 vs. Hₐ: p₁ - p₂ > 0.

(b)

The test statistic is defined as follows:

`z=\frac{\hat p_(1)-\hat p_(2)}{\sqrt{\hat P(1-\hat P)[(1)/(n_(1))+(1)/(n_(2))]}}`

The information provided is:

n₁ = 244

n₂ = 311

x₁ = 122

x₂ = 137

Compute the sample proportions and total proportions as follows:

`\hat p_(1)=(x_(1))/(n_(1))=(122)/(244)=0.50\\\\\hat p_(2)=(x_(2))/(n_(2))=(137)/(311)=0.44\\\\\hat P=(X_(1)+X_(2))/(n_(1)+n_(2))=(122+137)/(244+311)=0.47`

Compute the test statistic value as follows:

`z=\frac{\hat p_(1)-\hat p_(2)}{\sqrt{\hat P(1-\hat P)[(1)/(n_(1))+(1)/(n_(2))]}}`

`=\frac{0.50-0.44}{\sqrt{0.47(1-0.47)[(1)/(244)+(1)/(311)]}}\\\\=1.41`

The test statistic value is 1.41.

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level α = 0.05.

Compute the p-value as follows:

`p-value=P(Z>1.41)\\=1-P(Z<1.41)\\=1-0.92073\\=0.07927\\\approx 0.08`

*Use a z-table.

The p-value of the test is 0.08.

p-value = 0.08 > α = 0.05

The null hypothesis will not be rejected at 5% significance level.

Thus, there is not enough evidence to reject the null hypothesis.