 # Consider a random sample of size n = 31, with sample mean \bar{x}=45.2 and sample standard deviation s = 5.3. Compute 90%, 95%, and 99% confidence int boitshupoO 2021-05-05 Answered

Consider a random sample of size n = 31, with sample mean $\stackrel{―}{x}=45.2$ and sample standard deviation s = 5.3. Compute 90%, 95%, and 99% confidence intervals for \muμ using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.

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Given
$\stackrel{―}{x}=45.2$
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:

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