Question

A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second population revealed x2 to be 110. Use the 0.05 significance level to test the hypothesis.

Answer 1

a) For this case the value of the significanceis
`\alpha=0.05` and
`\alpha/2 =0.025`, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

`z_(\alpha/2) =1.96`

If the calculated statistic
`|z_(calc)| >1.96` we can reject the null hypothesis at 5% of significance

b) Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(170+110)/(200+150)=0.8`

c)
`z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)((1)/(200)+(1)/(150))}}=2.708`

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Explanation:

Data given and notation

`X_(1)=170` represent the number of people with the characteristic 1

`X_(2)=110` represent the number of people with the characteristic 2

`n_(1)=200` sample 1 selected

`n_(2)=150` sample 2 selected

`p_(1)=(170)/(200)=0.85` represent the proportion estimated for the sample 1

`p_(2)=(110)/(150)=0.733` represent the proportion estimated for the sample 2

`\hat p` represent the pooled estimate of p

z would represent the statistic (variable of interest)

`p_v` represent the value for the test (variable of interest)

`\alpha=0.05` significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis:
`p_(1) = p_(2)`

Alternative hypothesis:
`p_(1) \\eq p_(2)`

We need to apply a z test to compare proportions, and the statistic is given by:

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

a.State the decision rule.

For this case the value of the significanceis
`\alpha=0.05` and
`\alpha/2 =0.025`, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

`z_(\alpha/2) =1.96`

If the calculated statistic
`|z_(calc)| >1.96` we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(170+110)/(200+150)=0.8`

c. Compute the value of the test statistic.

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)((1)/(200)+(1)/(150))}}=2.708`

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.