Generating...                               quiz03_n8

  1. A consumer agency suspects that a pet food company may be underfilling packages for one of its brands. The package label states “ 800 grams net weight,” and the president of the company claims the average weight is greater than the amount stated. For a random sample of 12 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 744.13 grams and the sample standard deviation is $S$ = 57.821. Use the significance level $\alpha = 0.1$, and find the correct statement.

    $T = ( 744.13 - 800)/$ ( $\displaystyle {{28.9105}\over{\sqrt{3}}}$) = -3.347 is less than -1.363, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 800$. Thus, the evidence for underfilling is statistically not significant . $T = ( 744.13 - 800)/$ ( $\displaystyle {{57.821}\over{\sqrt{11}}}$) = -3.205 is less than -1.363, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 800$. Thus, the evidence for underfilling is statistically significant . $T = ( 744.13 - 800)/$ ( $\displaystyle {{28.9105}\over{\sqrt{3}}}$) = -3.347 is less than -1.363, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 800$. Thus, the evidence for underfilling is statistically significant . $T = ( 744.13 - 800)/$ ( $\displaystyle {{57.821}\over{\sqrt{11}}}$) = -3.205 is less than -1.363, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 800$. Thus, the evidence for underfilling is statistically not significant .

  2. A sample of 21 observations has a sample mean $\bar{X} = 28.38$ and a sample standard deviation $S = 14.62$. Find a 95% two-sided confidence interval for the population mean.

    $28.38 \pm ( 2.08) ( 14.62)/$ $\displaystyle \sqrt{21}$ $28.38 \pm ( 1.721) ( 14.62)/$ $\displaystyle \sqrt{21}$ $28.38 \pm ( 2.086) ( 14.62)/$ $\displaystyle 2\,\sqrt{5}$ $28.38 \pm ( 1.725) ( 14.62)/$ $\displaystyle \sqrt{21}$ $28.38 \pm ( 1.725) ( 14.62)/$ $\displaystyle 2\,\sqrt{5}$ $28.38 \pm ( 2.086) ( 14.62)/$ $\displaystyle \sqrt{21}$

  3. A machine is set to cut metal plates to a length of 80 mm. The length of a random sample of 25 metal plates have a sample mean of $\bar{X}$ = 80.081 mm and a sample standard deviation of $S$ = 0.176 mm. Is there any evidence that the machine is miscalibrated? Use the significance level $\alpha = 0.01$, and find the correct statement.

    $\vert T\vert = \vert 80.081 - 80\vert/$ (0.03519999999999999) = 2.301 is less than 2.797, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically significant . $\vert T\vert = \vert 80.081 - 80\vert/$ (0.03519999999999999) = 2.301 is less than 2.492, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically not significant . $\vert T\vert = \vert 80.081 - 80\vert/$ (0.03519999999999999) = 2.301 is less than 2.797, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically not significant .

  4. 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 $n$ = 15 bottles of the chemical solution is obtained and their densities are measured, and that the sample mean $\bar{X}$ = 1.113 and the sample standard deviation is $S$ = 0.178. Use the significance level $\alpha = 0.1$, and find the correct statement.

    Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{14}}}$) = 4.477 is greater than 1.345, the null hypothesis can be rejected. Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{14}}}$) = 4.477 is greater than 1.345, the null hypothesis cannot be rejected. Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{15}}}$) = 4.635 is greater than 1.761, the null hypothesis can be rejected. Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{15}}}$) = 4.635 is greater than 1.345, the null hypothesis can be rejected. Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{15}}}$) = 4.635 is greater than 1.761, the null hypothesis cannot be rejected. Since $T = ( 1.113 - 0.9)/$ ( $\displaystyle {{0.178}\over{\sqrt{15}}}$) = 4.635 is greater than 1.345, the null hypothesis cannot be rejected.

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


Department of Mathematics
Last modified: 2026-03-24