Consider a random sample of size n = 31, with sample mean

boitshupoO
2021-05-05
Answered

Consider a random sample of size n = 31, with sample mean

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2abehn

Answered 2021-05-06
Author has **88** answers

Given

$\stackrel{\u2015}{x}=45.2$

n=31

$\sigma =5.3$

Determine the t-value by looking in the row starting with degrees of freedom df=n-1=31-1=30 and in the column with c=90%/95%/99% in table 4:

$90\mathrm{\%}:{t}_{\frac{\alpha}{2}}=1.697$

$95\mathrm{\%}:{t}_{\frac{\alpha}{2}}=2.042$

$99\mathrm{\%}:{t}_{\frac{\alpha}{2}}=2.750$

The confidence interval is:

$\stackrel{\u2015}{x}-{t}_{\frac{\alpha}{2}}\dot{\{}\frac{s}{\sqrt{n}}\text{}\to \text{}\stackrel{\u2015}{x}+{t}_{\frac{\alpha}{2}}\dot{\frac{s}{\sqrt{n}}}$

Fill in the known values: NSL$45.2-{t}_{\frac{\alpha}{2}}\dot{\{}\frac{5.3}{\sqrt{31}}\text{}\to \text{}45.2+{t}_{\frac{\alpha}{2}}\dot{\{}\frac{5.3}{\sqrt{31}}$

Simplify:

90%:43.5846 to 46.8154

95%:43.2562 to 47.1438

99%:42.5823 to 47.8177

n=31

Determine the t-value by looking in the row starting with degrees of freedom df=n-1=31-1=30 and in the column with c=90%/95%/99% in table 4:

The confidence interval is:

Fill in the known values: NSL

Simplify:

90%:43.5846 to 46.8154

95%:43.2562 to 47.1438

99%:42.5823 to 47.8177

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