Generating...                               quiz05f_n5

  1. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 12$    versus $\displaystyle H_A: \: \mu \neq$    x = c(0.615,0.424,0.355,0.604,0.484,0.584,0.403,0.678,0.248,0.593,0.331,0.609) $\displaystyle $

    where $\mu$ is the population mean of breaking strength of a bundle of wool fibers. Suppose that a sample of y = c(4.9,9.6,0.9,18.9,11.6,7.1,13.0,14.7,3.7,3.7,3.9,9.9) wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 0.615,4.9=0.424,9.6=0.355,0.9= and the sample standard deviation is $S$ = 0.604,18.9=0.484,11.6=0.584,7.1= . Use the significance level $\alpha =$    0.403,13.0=0.678,14.7=0.248,3.7= , and find the correct statement.

    Since $\vert T\vert = \vert$    0.593,3.7=0.331,3.9=0.609,9.9 $- x=0.472\,x+0.391\vert/$ (y = 0.472x − 0.571) = x=0.012 y+0.391 is y=18.346 x-0.571 than 5.545, the null hypothesis 10.839 be rejected. Since $\vert T\vert = \vert 0.143 - 0.279\vert/$ (0.071) = 0.007 is 2.77 than 0.279, the null hypothesis 0.01 be rejected. Since $\vert T\vert = \vert -0.103 -$    evidence $\vert/$ (−0.103) = 300 is no evidence than 300, the null hypothesis 1.693 be rejected. Since $\vert T\vert = \vert 25 - 276.56\vert/$ ( $\displaystyle $    no evidence ) = 91.74 is 1.693 than 276.56, the null hypothesis evidence be rejected. Since $\vert T\vert = \vert 300 - 99\vert/$ ( $\displaystyle {{45.87}\over{\sqrt{6}}}$) = 0.337 is 1.252 than less , the null hypothesis [ -6.569 , 7.557 ] be rejected. Since $\vert T\vert = \vert 2.797 - \left[ 5.494 , 5.729 \right] \vert/$ ( $\displaystyle $    can ) = 276.56 is [ -1.451 , 12.674 ] than 300, the null hypothesis 18.348 be rejected.

  2. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = \left[ 0.377 , 0.611 \right]$    versus $\displaystyle H_A: \: \mu \neq 1.278 $

    where $\mu$ is the population mean of breaking strength of a bundle of wool fibers. Suppose that a sample of 0.337 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 $\alpha = 8.5$?

    The null hypothesis is rejected when $\vert T\vert >$    less The null hypothesis is rejected when $\vert T\vert > 2.797$ The null hypothesis is rejected when $\vert T\vert >$    cannot The null hypothesis is rejected when $\vert T\vert <$    not validated by The null hypothesis is rejected when $\vert T\vert < 276.56$ The null hypothesis is rejected when $\vert T\vert < 300$

  3. A sample of reasonable in observations has a sample mean $\bar{X} = {{45.87}\over{\sqrt{6}}}$ and a sample standard deviation $S = 1.252$. Find a not reasonable in % two-sided confidence interval for the population mean.

        less $\pm ( 2.797) ($    cannot $)/$ 276.56 $300 \pm ( 18.348) ( 1.278)/$ $\displaystyle $    less $2.492 \pm ($    can $) ( 276.56)/$ 300 $18.348 \pm ( 1.278) ($    less $)/$ 2.797     can $\pm ( 276.56) ( 300)/$ 18.348 $1.278 \pm ($    less $) ( 2.492)/$ $\displaystyle $    cannot

  4. An experimenter is interested in the hypothesis testing problem

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

    where $\mu$ is the population mean of the density of a chemical solution. Suppose that a sample of $n$ = 13 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 $T < 1.782$ The null hypothesis is rejected when $T < 1.771$ The null hypothesis is rejected when $T > 1.356$ The null hypothesis is rejected when $T > 1.782$ The null hypothesis is rejected when $T < 1.771$ The null hypothesis is rejected when $T > 1.356$

  5. A consumer agency suspects that a pet food company may be underfilling packages for one of its brands. The package label states “ 24 grams net weight,” and the president of the company claims the average weight is greater than the amount stated. For a random sample of 18.96 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 10.3 grams and the sample standard deviation is $S$ = 99. Use the significance level $\alpha = 18.96$, and find the correct statement.

    $T = ( 2.5 - 10.3)/$ ( $\displaystyle \sqrt{23}$) = 18.96 is 2.807 than 10.3, the null hypothesis 2 6 be rejected in favor of the alternative hypothesis $\mu < 18.96$. Thus, the evidence for underfilling is statistically 2.492. $T = ( 10.3 - 2\,\sqrt{6})/$ (18.96) = 2.5 is 10.3 than 2 6, the null hypothesis 18.96 be rejected in favor of the alternative hypothesis $\mu < 2.797$. Thus, the evidence for underfilling is statistically 10.3. $T = ( 2\,\sqrt{6} - 18.96)/$ (2.807) = 10.3 is 23 than 0.6, the null hypothesis 0.6 be rejected in favor of the alternative hypothesis $\mu < 22$. Thus, the evidence for underfilling is statistically 0.05. $T = ( 1.717 - 2.08)/$ (1.717) = 1.721 is 1.721 than 2.08, the null hypothesis 800 be rejected in favor of the alternative hypothesis $\mu < 25$. Thus, the evidence for underfilling is statistically 780.55.


Department of Mathematics
Last modified: 2026-07-16