1. A sample of 11 observations has a sample mean $\bar{X} = 29.0$ and a sample standard deviation $S = 20.11$. Find a 95% two-sided confidence interval for the population mean.

    $29.0 \pm ( 1.796) ( 20.11)/$ $\displaystyle \sqrt{11}$ $29.0 \pm ( 2.228) ( 20.11)/$ $\displaystyle \sqrt{11}$ $29.0 \pm ( 1.812) ( 20.11)/$ $\displaystyle \sqrt{11}$ $29.0 \pm ( 2.228) ( 20.11)/$ $\displaystyle \sqrt{10}$ $29.0 \pm ( 1.812) ( 20.11)/$ $\displaystyle \sqrt{10}$ $29.0 \pm ( 2.201) ( 20.11)/$ $\displaystyle \sqrt{11}$

  2. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 300$    versus $\displaystyle H_A: \: \mu \neq 300 $

    where $\mu$ is the population mean of breaking strength of a bundle of wool fibers. Suppose that a sample of 13 wool fiber bundles is obtained and their breaking strengths are measured. For what values of the $t$-statistic $T$ does the experimenter reject the null hypothesis with significance level $\alpha = 0.1$?

    The null hypothesis is rejected when $\vert T\vert > 1.356$ The null hypothesis is rejected when $\vert T\vert < 1.356$ The null hypothesis is rejected when $\vert T\vert > 1.782$ The null hypothesis is rejected when $\vert T\vert < 1.771$ The null hypothesis is rejected when $\vert T\vert < 1.782$ The null hypothesis is rejected when $\vert T\vert > 1.771$

  3. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 0.9$    versus $\displaystyle H_A: \: \mu > 0.9 $

    where $\mu$ is the population mean of the density of a chemical solution. Suppose that a sample of $n$ = 20 bottles of the chemical solution is obtained and their densities are measured. For what values of the $t$-statistic $T$ does the experimenter reject the null hypothesis with significance level $\alpha = 0.1$?

    The null hypothesis is rejected when $T < 1.328$ The null hypothesis is rejected when $T > 1.729$ The null hypothesis is rejected when $T < 1.325$ The null hypothesis is rejected when $T < 1.729$ The null hypothesis is rejected when $T > 1.328$ The null hypothesis is rejected when $T > 1.325$

  4. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 0.6$    versus $\displaystyle H_A: \: \mu > 0.6 $

    where $\mu$ is the population mean of the density of a chemical solution. Suppose that a sample of $n$ = 20 bottles of the chemical solution is obtained and their densities are measured, and that the sample mean $\bar{X}$ = 0.538 and the sample standard deviation is $S$ = 0.478. Use the significance level $\alpha = 0.01$, and find the correct statement.

    Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.239}\over{\sqrt{5}}}$) = -0.58 is less than 2.539, the null hypothesis can be rejected. Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.478}\over{\sqrt{19}}}$) = -0.565 is less than 2.539, the null hypothesis can be rejected. Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.239}\over{\sqrt{5}}}$) = -0.58 is less than 2.539, the null hypothesis cannot be rejected. Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.478}\over{\sqrt{19}}}$) = -0.565 is less than 2.539, the null hypothesis cannot be rejected. Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.239}\over{\sqrt{5}}}$) = -0.58 is less than 2.861, the null hypothesis can be rejected. Since $T = ( 0.538 - 0.6)/$ ( $\displaystyle {{0.239}\over{\sqrt{5}}}$) = -0.58 is less than 2.861, the null hypothesis cannot be rejected.

  5. A machine is set to cut metal plates to a length of 40 mm. The length of a random sample of 13 metal plates have a sample mean of $\bar{X}$ = 39.939 mm and a sample standard deviation of $S$ = 0.25 mm. Is there any evidence that the machine is miscalibrated? Use the significance level $\alpha = 0.1$, and find the correct statement.

    $\vert T\vert = \vert 39.939 - 40\vert/$ ( $\displaystyle {{0.25}\over{\sqrt{13}}}$) = 0.88 is less than 1.782, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically significant . $\vert T\vert = \vert 39.939 - 40\vert/$ ( $\displaystyle {{0.25}\over{\sqrt{13}}}$) = 0.88 is less than 1.356, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically not significant . $\vert T\vert = \vert 39.939 - 40\vert/$ ( $\displaystyle {{0.25}\over{\sqrt{13}}}$) = 0.88 is less than 1.782, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically not significant .


Department of Mathematics
Last modified: 2025-10-12