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elena-s [515]
3 years ago
10

Round 2.16993 to the nearest thousandth​

Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

2.170

Step-by-step explanation:

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According to Nielsen Media Research. of all the U.S. households that owned at least one television set, 83% had two or more sets
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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.

<u>Assumptions:</u>

The sample size (<em>n</em>) 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 <em>p</em> and standard deviation \sqrt{\frac{p(1-p)}{n} } .

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

<u>The hypothesis is:</u>

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

<u>Decision Rule:</u>

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

\\ Z\leq -1.645\\

**Use the <em>z</em> table for the critical values.

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

<u>Test statistic value:</u>

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.

<u>Conclusion:</u>

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%.

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