Question

(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)

Answer 1

The margin of error (E)
`=0.49`

Explanation:

Given,

mean
`\bar{x}=6.88`

standard deviation
`\sigma=1.90`

sample size
`n=41`

`90\% \quad \text{confidence interval}`

Critical value
`=(z_\alpha)/(2)\quad \quad\quad[\because\;\alpha=1-\text{confidence interval}]`

`=1-0.90`

`\Rightarrow \alpha=0.1`

Critical value
`=(z_(0.1))/(2)`

`=z_(0.05)`

`=1.645`

The margin of error (E)
`=(z_\alpha)/(2) *(\sigma)/(√(n) )`

`=1.645*(1.90)/(√(41) )`

`=0.49`

Hence, The margin of error (E)
`=0.49`

Complete question is attached in below.