Generating...                               quiz07c_n10

  1. A sample of 12 observations has a sample mean $\bar{X} = 33.25$ and a sample standard deviation $S = 12.45$. Find a 99% two-sided confidence interval for the population mean.

    $33.25 \pm ( 3.106) ( 12.45)/$ $\displaystyle 2\,\sqrt{3}$ $33.25 \pm ( 2.681) ( 12.45)/$ $\displaystyle 2\,\sqrt{3}$ $33.25 \pm ( 3.106) ( 12.45)/$ $\displaystyle \sqrt{11}$ $33.25 \pm ( 2.718) ( 12.45)/$ $\displaystyle \sqrt{11}$ $33.25 \pm ( 3.055) ( 12.45)/$ $\displaystyle 2\,\sqrt{3}$ $33.25 \pm ( 2.718) ( 12.45)/$ $\displaystyle 2\,\sqrt{3}$

  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 16 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 $\alpha = 0.05$?

    The null hypothesis is rejected when $\vert T\vert < 2.131$ The null hypothesis is rejected when $\vert T\vert > 2.12$ The null hypothesis is rejected when $\vert T\vert > 2.131$ The null hypothesis is rejected when $\vert T\vert < 1.753$ The null hypothesis is rejected when $\vert T\vert > 1.753$ The null hypothesis is rejected when $\vert T\vert < 2.12$

  3. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 200$    versus $\displaystyle H_A: \: \mu \neq 200 $

    where $\mu$ is the population mean of breaking strength of a bundle of wool fibers. Suppose that a sample of 24 wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 141.88 and the sample standard deviation is $S$ = 105.2. Use the significance level $\alpha = 0.05$, and find the correct statement.

    Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{52.6}\over{\sqrt{6}}}$) = 2.707 is greater than 2.069, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{52.6}\over{\sqrt{6}}}$) = 2.707 is greater than 1.714, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{105.2}\over{\sqrt{23}}}$) = 2.65 is greater than 2.069, the null hypothesis can be rejected. Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{52.6}\over{\sqrt{6}}}$) = 2.707 is greater than 2.069, the null hypothesis can be rejected. Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{105.2}\over{\sqrt{23}}}$) = 2.65 is greater than 2.069, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 141.88 - 200\vert/$ ( $\displaystyle {{52.6}\over{\sqrt{6}}}$) = 2.707 is greater than 1.714, the null hypothesis can be rejected.

  4. A machine is set to cut metal plates to a length of 40 mm. The length of a random sample of 29 metal plates have a sample mean of $\bar{X}$ = 40.101 mm and a sample standard deviation of $S$ = 0.171 mm. Is there any evidence that the machine is miscalibrated? Use the significance level $\alpha = 0.01$, and find the correct statement.

    $\vert T\vert = \vert 40.101 - 40\vert/$ ( $\displaystyle {{0.171}\over{\sqrt{29}}}$) = 3.181 is greater than 2.763, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically not significant . $\vert T\vert = \vert 40.101 - 40\vert/$ ( $\displaystyle {{0.171}\over{\sqrt{29}}}$) = 3.181 is greater than 2.763, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically significant . $\vert T\vert = \vert 40.101 - 40\vert/$ ( $\displaystyle {{0.171}\over{\sqrt{29}}}$) = 3.181 is greater than 2.467, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 40$. Thus, the evidence of miscalibration is statistically significant .

  5. 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 $n$ = 11 bottles of the chemical solution is obtained and their densities are measured, and that the sample mean $\bar{X}$ = 0.835 and the sample standard deviation is $S$ = 0.386. Use the significance level $\alpha = 0.01$, and find the correct statement.

    Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{11}}}$) = 2.019 is less than 2.764, the null hypothesis can be rejected. Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{10}}}$) = 1.925 is less than 2.764, the null hypothesis cannot be rejected. Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{11}}}$) = 2.019 is less than 3.169, the null hypothesis can be rejected. Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{10}}}$) = 1.925 is less than 2.764, the null hypothesis can be rejected. Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{11}}}$) = 2.019 is less than 3.169, the null hypothesis cannot be rejected. Since $T = ( 0.835 - 0.6)/$ ( $\displaystyle {{0.386}\over{\sqrt{11}}}$) = 2.019 is less than 2.764, the null hypothesis cannot be rejected.


Department of Mathematics
Last modified: 2025-06-19