Generating...                               quiz03_n22

  1. 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 17 wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 300.41 and the sample standard deviation is $S$ = 37.07. Use the significance level $\alpha = 0.1$, and find the correct statement.

    Since $\vert T\vert = \vert 300.41 - 300\vert/$ ( $\displaystyle {{37.07}\over{\sqrt{17}}}$) = 0.046 is less than 1.337, the null hypothesis can be rejected. Since $\vert T\vert = \vert 300.41 - 300\vert/$ (9.2675) = 0.044 is less than 1.746, the null hypothesis can be rejected. Since $\vert T\vert = \vert 300.41 - 300\vert/$ ( $\displaystyle {{37.07}\over{\sqrt{17}}}$) = 0.046 is less than 1.746, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 300.41 - 300\vert/$ ( $\displaystyle {{37.07}\over{\sqrt{17}}}$) = 0.046 is less than 1.746, the null hypothesis can be rejected. Since $\vert T\vert = \vert 300.41 - 300\vert/$ ( $\displaystyle {{37.07}\over{\sqrt{17}}}$) = 0.046 is less than 1.337, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 300.41 - 300\vert/$ (9.2675) = 0.044 is less than 1.746, the null hypothesis cannot be rejected.

  2. 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 $n$ = 16 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 $\alpha = 0.05$?

    The null hypothesis is rejected when $T > 1.746$ The null hypothesis is rejected when $T < 1.753$ The null hypothesis is rejected when $T > 1.753$ The null hypothesis is rejected when $T < 1.746$ The null hypothesis is rejected when $T < 2.131$ The null hypothesis is rejected when $T > 2.131$

  3. 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 12 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 $\vert T\vert < 3.055$ The null hypothesis is rejected when $\vert T\vert < 2.718$ The null hypothesis is rejected when $\vert T\vert > 3.106$ The null hypothesis is rejected when $\vert T\vert > 3.055$ The null hypothesis is rejected when $\vert T\vert > 2.718$ The null hypothesis is rejected when $\vert T\vert < 3.106$

  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 $n$ = 19 bottles of the chemical solution is obtained and their densities are measured, and that the sample mean $\bar{X}$ = 1.291 and the sample standard deviation is $S$ = 0.293. Use the significance level $\alpha = 0.01$, and find the correct statement.

    Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.293}\over{\sqrt{19}}}$) = 5.817 is greater than 2.552, the null hypothesis cannot be rejected. Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.293}\over{\sqrt{19}}}$) = 5.817 is greater than 2.878, the null hypothesis can be rejected. Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.293}\over{\sqrt{19}}}$) = 5.817 is greater than 2.552, the null hypothesis can be rejected. Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.293}\over{\sqrt{19}}}$) = 5.817 is greater than 2.878, the null hypothesis cannot be rejected. Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.09766666666666665}\over{\sqrt{2}}}$) = 5.662 is greater than 2.552, the null hypothesis can be rejected. Since $T = ( 1.291 - 0.9)/$ ( $\displaystyle {{0.09766666666666665}\over{\sqrt{2}}}$) = 5.662 is greater than 2.552, the null hypothesis cannot be rejected.

  5. 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 19 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 1205.142 grams and the sample standard deviation is $S$ = 25.326. Use the significance level $\alpha = 0.05$, and find the correct statement.

    $T = ( 1205.142 - 1200)/$ ( $\displaystyle {{8.442}\over{\sqrt{2}}}$) = 0.861 is greater than -1.734, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically significant . $T = ( 1205.142 - 1200)/$ ( $\displaystyle {{25.326}\over{\sqrt{19}}}$) = 0.885 is greater than -1.734, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically not significant . $T = ( 1205.142 - 1200)/$ ( $\displaystyle {{8.442}\over{\sqrt{2}}}$) = 0.861 is greater than -1.734, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically not significant . $T = ( 1205.142 - 1200)/$ ( $\displaystyle {{25.326}\over{\sqrt{19}}}$) = 0.885 is greater than -1.734, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1200$. Thus, the evidence for underfilling is statistically significant .


Department of Mathematics
Last modified: 2025-09-14