Generating...                               quiz03_n2

  1. A consumer agency suspects that a pet food company may be underfilling packages for one of its brands. The package label states “ 1200 grams net weight,” and the president of the company claims the average weight is greater than the amount stated. For a random sample of 20 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 1160.737 grams and the sample standard deviation is $S$ = 69.106. Use the significance level $\alpha = 0.1$, and find the correct statement.

    $T = ( 1160.737 - 1200)/$ ( $\displaystyle {{69.106}\over{\sqrt{19}}}$) = -2.477 is less than -1.328, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically significant . $T = ( 1160.737 - 1200)/$ ( $\displaystyle {{69.106}\over{\sqrt{19}}}$) = -2.477 is less than -1.328, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically not significant . $T = ( 1160.737 - 1200)/$ ( $\displaystyle {{34.553}\over{\sqrt{5}}}$) = -2.541 is less than -1.328, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically significant . $T = ( 1160.737 - 1200)/$ ( $\displaystyle {{34.553}\over{\sqrt{5}}}$) = -2.541 is less than -1.328, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically not significant .

  2. 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 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 $\alpha = 0.01$?

    The null hypothesis is rejected when $\vert T\vert < 2.947$ The null hypothesis is rejected when $\vert T\vert < 2.921$ The null hypothesis is rejected when $\vert T\vert > 2.947$ The null hypothesis is rejected when $\vert T\vert > 2.921$ The null hypothesis is rejected when $\vert T\vert < 2.602$ The null hypothesis is rejected when $\vert T\vert > 2.602$

  3. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 400$    versus $\displaystyle H_A: \: \mu \neq 400 $

    where $\mu$ is the population mean of breaking strength of a bundle of wool fibers. Suppose that a sample of 12 wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 409.42 and the sample standard deviation is $S$ = 66.04. Use the significance level $\alpha = 0.01$, and find the correct statement.

    Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{33.02}\over{\sqrt{3}}}$) = 0.494 is less than 2.718, the null hypothesis can be rejected. Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{33.02}\over{\sqrt{3}}}$) = 0.494 is less than 2.718, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{66.04}\over{\sqrt{11}}}$) = 0.473 is less than 3.106, the null hypothesis can be rejected. Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{66.04}\over{\sqrt{11}}}$) = 0.473 is less than 3.106, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{33.02}\over{\sqrt{3}}}$) = 0.494 is less than 3.106, the null hypothesis can be rejected. Since $\vert T\vert = \vert 409.42 - 400\vert/$ ( $\displaystyle {{33.02}\over{\sqrt{3}}}$) = 0.494 is less than 3.106, the null hypothesis cannot be rejected.

  4. A machine is set to cut metal plates to a length of 80 mm. The length of a random sample of 11 metal plates have a sample mean of $\bar{X}$ = 79.891 mm and a sample standard deviation of $S$ = 0.269 mm. Is there any evidence that the machine is miscalibrated? Use the significance level $\alpha = 0.05$, and find the correct statement.

    $\vert T\vert = \vert 79.891 - 80\vert/$ ( $\displaystyle {{0.269}\over{\sqrt{11}}}$) = 1.344 is less than 2.228, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically significant . $\vert T\vert = \vert 79.891 - 80\vert/$ ( $\displaystyle {{0.269}\over{\sqrt{11}}}$) = 1.344 is less than 1.812, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically not significant . $\vert T\vert = \vert 79.891 - 80\vert/$ ( $\displaystyle {{0.269}\over{\sqrt{11}}}$) = 1.344 is less than 2.228, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 80$. Thus, the evidence of miscalibration is statistically not significant .

  5. A sample of 17 observations has a sample mean $\bar{X} = 29.35$ and a sample standard deviation $S = 14.83$. Find a 99% two-sided confidence interval for the population mean.

    $29.35 \pm ( 2.583) ( 14.83)/$ $\displaystyle \sqrt{17}$ $29.35 \pm ( 2.583) ( 14.83)/$ 4 $29.35 \pm ( 2.921) ( 14.83)/$ 4 $29.35 \pm ( 2.921) ( 14.83)/$ $\displaystyle \sqrt{17}$ $29.35 \pm ( 2.567) ( 14.83)/$ $\displaystyle \sqrt{17}$ $29.35 \pm ( 2.898) ( 14.83)/$ $\displaystyle \sqrt{17}$


Department of Mathematics
Last modified: 2025-06-01