Answer:
Test statistic = 1.3471
P-value = 0.1993
Accept the null hypothesis.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4
Sample mean, [tex]\bar{x}[/tex] = 4.8
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation, s = 2.3
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 4\\H_A: \mu \neq 4[/tex]
We use two-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{4.8 - 4}{\frac{2.3}{\sqrt{15}} } = 1.3471[/tex]
Now, we calculate the p-value.
P-value = 0.1993
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.
