Question

NO LINKS!! Please help me with this problem 1b​

Answer 1

`\textsf{b)} \quad 15.8 \pm 0.17`

Step-by-step explanation:

Given values:

• Sample size n = 200
• Population standard deviation σ = 1.2
• Sample mean
  `\overline{x}` = 15.8

`\boxed{\begin{minipage}{7 cm}\underline{Standard error of the mean}\\\\$SE=(\sigma)/(√(n))$\\\\where:\\\phantom{ww}$\bullet$ $\sigma$ is the population standard deviation. \\ \phantom{ww}$\bullet$ $n$ is the sample size.\\\end{minipage}}`

Substitute the given values into the standard error formula:

`\implies SE=(1.2)/(√(200))=(3√(2))/(50)`

The critical value for a 95% confidence level using normal distribution is:

`z=1.9600\;\;\sf (4\;d.p.)`

`\boxed{\begin{minipage}{6.7 cm}\underline{Margin of error}\\\\$ME=z \cdot SE$\\\\where:\\\phantom{ww}$\bullet$ $z$ is the critical value. \\ \phantom{ww}$\bullet$ $SE$ is the standard error of the mean.\\\end{minipage}}`

Substitute the given values into the margin of error formula:

`\implies ME=1.9600 \cdot (3√(2))/(50)`

`\implies ME=0.17\;\; \sf (2\;d.p.)`

Therefore, an estimate for the size of the population mean, including margin or error is:

`\implies \mu=\overline{x}\pm ME`

`\implies \mu=15.8 \pm 0.17`

Answer 2

Answer: Choice B)
`\boldsymbol{15.8 \ \pm \ 0.17}`

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Step-by-step explanation:

sigma = 1.2 = population standard deviation

n = 200 = sample size

Your teacher doesn't mention the confidence level, so I'll assume we go for the default 95%.

At 95% confidence, the z critical value is roughly z = 1.96

The margin of error (E) is calculated like so:

E = z*sigma/sqrt(n)

E = 1.96*1.2/sqrt(200)

E = 0.1663 approximately

That rounds to 0.17

Therefore the estimate for the population mean in the format
`\overline{\text{x}} \ \pm \ E` is
`\boldsymbol{15.8 \ \pm \ 0.17}` which is the final answer.

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This is the shorthand way of saying the mean mu is between
`\overline{\text{x}} \ - \ E = 15.8 - 0.17 = 15.63` and
`\overline{\text{x}} \ + \ E = 15.8 +0.17 = 15.97`

In other words,

`15.63 < \mu < 15.97` at 95% confidence.