Question

According to Nielsen Media Research. of all the U.S. households that owned at least one television set, 83% had two or more sets. A local cable company canvassing the town to promote a new cable service found that of the 300 households visited, 240 had tow or more television sets. At α= .05 proportion is less than the one in the report.

Answer 1

The proportion of U.S. households that owned two or more televisions is 83%.

Explanation:

To determine whether the proportions of U.S. households that owned two or more televisions is less than 83% or not let us perform a hypothesis test for single proportion.

Assumptions:

The sample size (n) selected by the local cable company is 300 which is quite large. Then according to the Central limit theorem the sampling distribution of sample proportion follows a normal distribution with mean p and standard deviation
`\sqrt{(p(1-p))/(n) }` .

Since the sampling distribution of sample proportions follows a normal distribution use the z-test for one proportion to perform the test.

The hypothesis is:

`H_(0)` : The proportion of U.S. households that owned two or more televisions is 83%, i.e.
`p=0.83`

`H_(1)` :The proportion of U.S. households that owned two or more televisions is less than 83%, i.e.
`p< 0.83`

Decision Rule:

At the level of significance α = 0.05 the critical region for a one-tailed z-test is:

`\\ Z\leq -1.645\\`

**Use the z table for the critical values.

So, if
`\\ Z\leq -1.645\\` the null hypothesis will be rejected.

Test statistic value:

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}`

Here
`\hat{p}` is the sample proportion.

Compute the value of
`\hat{p}` as follows:

`\hat{p}=(X)/(n) \\=(240)/(300)\\ =0.80`

Now compute the value of the test statistic as follows:

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}\\=\frac{0.80-0.83}{\sqrt{(0.83*(1-0.83))/(300) } } \\=-1.383`

The test statistic is -1.383 which is more than -1.645.

Thus, the test statistic lies in the acceptance region.

Hence we fail to reject the null hypothesis.

Conclusion:

At 0.05 level of significance we fail to reject the null hypothesis stating that the proportion of U.S. households that owned two or more televisions is 83%.