Question

A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation of $0.24. It was felt that breakfasts on the East Coast were higher than $9.33. A sample of 81 business managers on the East Coast had an average breakfast cost of $9.41. At α=0.05, what is the test value?

Answer 1

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(9.41-9.33)/((0.24)/(√(81)))=3`

Explanation:

Information given

`\bar X=9.41` represent the sample mean

`s=0.24` represent the sample standard deviation

`n=81` sample size

`\mu_o =9.33` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 9.33`

Alternative hypothesis:
`\mu > 9.33`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(9.41-9.33)/((0.24)/(√(81)))=3`