Question

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

Bivariate numerical data
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asked 2021-05-05

Consider a random sample of size n = 31, with sample mean \(\displaystyle\overline{{{x}}}={45.2}\) 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.

Answers (1)

2021-05-06
Given
\(\displaystyle\overline{{x}}={45.2}\)
n=31
\(\displaystyle\sigma={5.3}\)
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:
\(\displaystyle{90}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={1.697}\)
\(\displaystyle{95}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.042}\)
\(\displaystyle{99}\%:{t}_{{{\frac{{\alpha}}{{{2}}}}}}={2.750}\)
The confidence interval is:
\(\displaystyle\overline{{x}}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\ \to\ \overline{{x}}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{{\frac{{{s}}}{{\sqrt{{{n}}}}}}}}\)
Fill in the known values: NSL \(\displaystyle{45.2}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\ \to\ {45.2}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\dot{{\lbrace}}{\frac{{{5.3}}}{{\sqrt{{{31}}}}}}\)
Simplify:
90%:43.5846 to 46.8154
95%:43.2562 to 47.1438
99%:42.5823 to 47.8177
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