Question

question 18
a recent publication claims that the average person spends 96 minutes per day on tiktok. a sociologist believes that adults in the baby boomer age range, spend less time on tiktok than the reported 96 minutes per day.
a random sample of 17 adults in the baby boomer age found the mean time on tiktok to be 87.841 minutes with a standard deviation of 11.2079 minutes. assume that the times were normally distributed.
test the sociologists claim using a 5% level of significance.
the hypotheses are:
$h_0: \mu = 96$
$h_a: \mu < 96$
based on the hypotheses, find the following:
a. t-score =
(round your answer to 4 decimal places.)
enter an integer or decimal number more..
b. p-value=
(round your answer to 4 decimal places.)

Explanation:

Step1: Identify t - test formula

For a one - sample t - test, the formula for the t - score is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.
We are given that \(\bar{x} = 87.841\), \(\mu=96\), \(s = 11.2079\), and \(n = 17\).

Step2: Calculate the t - score

First, calculate the numerator: \(\bar{x}-\mu=87.841 - 96=- 8.159\)
Then, calculate the denominator: \(s/\sqrt{n}=\frac{11.2079}{\sqrt{17}}\approx\frac{11.2079}{4.1231}\approx2.718\)
Now, calculate the t - score: \(t=\frac{- 8.159}{2.718}\approx - 3.002\) (rounded to 4 decimal places)

Step3: Calculate the p - value

We have a left - tailed test (since \(H_A:\mu<96\)) with \(n - 1=16\) degrees of freedom.
Using a t - distribution table or a statistical software, the p - value corresponding to \(t=-3.002\) and \(df = 16\) for a left - tailed test.
Using a t - calculator, the p - value is approximately \(0.0043\) (rounded to 4 decimal places)

Answer:

a. t - score \(\approx - 3.0020\)
b. p - value \(\approx0.0043\)