Math  /  Data & Statistics

QuestionA random sample of 10 independent healthy people showed the body temperatures given below (in degrees Fahrenheit). Test the hypothesis that the population mean is not 98.6F98.6^{\circ} \mathrm{F}, using a significance level of 0.05 . 98.598.499.096.198.898.797.299.298.997.5\begin{array}{llllllllll} 98.5 & 98.4 & 99.0 & 96.1 & 98.8 & 98.7 & 97.2 & 99.2 & 98.9 & 97.5 \end{array} answer below. A. H0:μ98.6H_{0}: \mu \neq 98.6 B. H0:μ=98.6H_{0}: \mu=98.6 C. H0:μ=98.6H_{0}: \mu=98.6 Ha:μ=98.6H_{a}: \mu=98.6 Ha:μ<98.6H_{a}: \mu<98.6 Ha:μ>98.6\mathrm{H}_{\mathrm{a}}: \mu>98.6 D. H0:μ<98.6H_{0}: \mu<98.6 E. H0:μ>98.6\mathrm{H}_{0}: \mu>98.6 F. H0:μ=98.6H_{0}: \mu=98.6 Ha:μ=98.6H_{a}: \mu=98.6 Ha:μ=98.6H_{a}: \mu=98.6 Ha:μ98.6H_{a}: \mu \neq 98.6
Find the test statistic. t=\mathrm{t}= \square (Round to two decimal places as needed.) Find the pp-value. p -value == \square (Round to three decimal places as needed.)

Studdy Solution
Make a decision based on the p-value:
Since the p-value 0.2070.207 is greater than the significance level α=0.05\alpha = 0.05, we fail to reject the null hypothesis.
The test statistic is:
t=1.36t = -1.36
The p-value is:
p-value=0.207p\text{-value} = 0.207

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