Question

A study is performed in a large southern town to determine whether the average weekly grocery bill per four-person family in the town is significantly different from the national average. The national average grocery bill is $100 (standard deviation unknown). A random sample of 25 families is taken in the town, and the mean is found to be 103.23 with a standard deviation of 8.82.

For each item below, please show your work and/or explain your reasoning to receive full credit.
a). State the null and alternative hypotheses for this situation.
b). Assuming a 0.05 level of significance, compute the appropriate test statistic.
c). Find the critical t value. Can you reject the null hypothesis? Explain/justify your response using numbers and words

Answer 1

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

Answer 2

The average weekly grocery bill per four-person family in the town is significantly different from the national average.

Explanation:

We are given the following in the question:

Population mean, μ = $100

Sample mean,
`\bar{x}` = $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

`H_(0): \mu = 100\text{ dollars}\\H_A: \mu \\eq 100\text{ dollars}`

We use Two-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(103.23- 100)/((8.82)/(√(25)) ) = 1.831`

Now,

`t_(critical) \text{ at 0.05 level of significance, 24 degree of freedom } = \pm 2.063`

Since,

`t_(stat) < t_(critical)`

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.