Question

According to a survey, the average American person watches TV for 3 hours per week. To test if the amount of TV in New York City is less than the national average, a researcher decides to do a hypothesis test, at a 1% significance level. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours. H0: μ≤3; Ha: μ>3. α=0.01 (significance level) What is the test statistic (t-value) of this one-mean hypothesis test (with σ unknown)?

Answer 1

Answer:

Test statistic (t-value) of this one-mean hypothesis test is -2.422.

Step-by-step explanation:

We are given that according to a survey, the average American person watches TV for 3 hours per week. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours.

We have to test if the amount of TV in New York City is less than the national average.

Let Null Hypothesis, $H_0$ : $\mu$ $\leq$ 3  {means that the amount of TV in New York City is less than the national average}

Alternate Hypothesis, $H_1$ : $\mu$ > 3   {means that the amount of TV in New York City is more than the national average}

The test statistics that will be used here is One-sample t-test statistics;

        T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean time = 2.5 hours per week

             $s$ = sample standard deviation = 0.09 hours

             n = sample size = 19

So, test statistics = $\frac{2.5 - 3}{\frac{0.9}{\sqrt{19} } }$ ~ $t_1_8$

                              = -2.422

Therefore, the test statistic (t-value) of this one-mean hypothesis test (with σ unknown) is -2.422.