Question

According to the U.S. Energy Information Administration the average number of televisions per household in the United States was 2.3. A college student claims the average number of TV’s per household in the United States is different. He obtains a random sample of 73 households and finds the mean number of TV’s to be 2.1 with a standard deviation of 0.84. Test the student’s claim at the 0.01 significance level.

Answer 1

Answer with explanation:

Let $\mu$ be the average number of televisions per household in the United States .

As per given ,

$H_0:\mu =2.3\\\\ H_a:\mu\neq2.3$

Since $H_a$ is two-tailed and population standard deviation is unknown, so the test is two-tailed t-test.

For sample : Sample size : n= 73, sample mean: $\overline{x}$ = 2.1, sample standard deviation : s= 0.84.

$t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$t=\dfrac{2.1-2.3}{\dfrac{0.84}{\sqrt{73}}}\\\\ t=-2.034$

T-critical value for degree of freedom n-1 = 73-1=72 and 0.01 significance level is 2.646 . [By students' t-distribution table]

Since, $|2.034|$ i.e. $|T_{cal}|$

This means we cannot reject null hypothesis.

We conclude that the average number of televisions per household in the United States is 2.3 at the 0.01 significance level.