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

Answer:

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

Step-by-step 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 $\sqrt{\frac{p(1-p)}{n} }$ .

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:

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

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

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

$\hat{p}=\frac{X}{n} \\=\frac{240}{300}\\ =0.80$

Now compute the value of the test statistic as follows:

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

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%.