Question

9. Henri and Terence drew samples of size 50 from a mystery bag. Henri drew 42 red chips, and Terence drew 40 red chips. Find the margins of error for each student.

Answer 1

The margins of error for Henri is 0.104.

The margins of error for Terence is 0.113.

Explanation:

Consider the provided information.

Henri and Terence drew samples of size 50 from a mystery bag.

Henri drew 42 red chips,

`\hat p=(42)/(50)=0.84`

The formula to calculate margins of error is:
`2\sqrt{(\hat p(1-\hat p))/(n)}`

Substitute n = 50 and
`\hat p=0.84` in above formula.

Margin of error for Henri =
`2\sqrt{(0.84(1-0.84))/(50)}`

Margin of error for Henri =
`2\sqrt{(0.84(0.16))/(50)}`

Margin of error for Henri ≈ 0.104

Hence, the margins of error for Henri is 0.104.

Terence drew 40 red chips,

`\hat p=(40)/(50)=0.84`

The formula to calculate margins of error is:
`2\sqrt{(\hat p(1-\hat p))/(n)}`

Substitute n = 50 and
`\hat p=0.80` in above formula.

Margin of error for Terence =
`2\sqrt{(0.80(1-0.80))/(50)}`

Margin of error for Terence =
`2\sqrt{(0.8(0.2))/(50)}`

Margin of error for Terence ≈ 0.113

Hence, the margins of error for Terence is 0.113.

Answer 2

Henri’s margin of error = 0.104

Terence’s margin of error = 0.113

Explanation:

It is given that Henri and Terence drew samples of size 50 from a mystery bag.

Formula for margin of error

`M.E.=2\sqrt{(p(1-p))/(n)}` ... (1)

where, p is the proportion.

Henri drew 42 red chips out of 50.

Proportion of getting red chip =
`(42)/(50)=0.84`

Substitute p=0.84 and n=50 in equation (1).

`M.E.=2\sqrt{((0.84)(1-0.84))/(50)}`

`M.E.=2\sqrt{((0.84)(0.16))/(50)}`

`M.E.=0.10369`

`M.E.\approx 0.104`

Henri’s margin of error is 0.104.

Terence drew 40 red chips.

Proportion of getting red chip =
`(40)/(50)=0.8`

Substitute p=0.8 and n=50 in equation (1).

`M.E.=2\sqrt{((0.8)(1-0.8))/(50)}`

`M.E.=2\sqrt{((0.8)(0.2))/(50)}`

`M.E.=0.113137`

`M.E.\approx 0.113`

Terence’s estimated margin of error is 0.113.