Answer:
The average weekly grocery bill per four-person family in the town is significantly different from the national average.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $100
Sample mean, [tex]\bar{x}[/tex] = $103.23
Sample size, n = 25
Alpha, α = 0.05
Sample standard deviation, s = $8.82
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 100\text{ dollars}\\H_A: \mu \neq 100\text{ dollars}[/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{103.23- 100}{\frac{8.82}{\sqrt{25}} } = 1.831[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = \pm 2.063[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We accept the null hypothesis.
We accept the null hypothesis and determine that the average weekly grocery bill per four-person family in the town is significantly same to the national average.
Answer:
From the given problem statement
Average=$100
mean=103.23
standard deviation=8.82
Significance level=0.05
confidence level=1-0.05=99.95
The null hypothesis is given below
H0=100
H0>103.23
