Generating...                               quiz03_n5

  1. A sample of 24 observations has a sample mean $\bar{X} = 21.79$ and a sample standard deviation $S = 8.22$. Find a 99% two-sided confidence interval for the population mean.

    $21.79 \pm ( 2.807) ( 8.22)/$ $\displaystyle 2\,\sqrt{6}$ $21.79 \pm ( 2.797) ( 8.22)/$ $\displaystyle 2\,\sqrt{6}$ $21.79 \pm ( 2.807) ( 8.22)/$ $\displaystyle \sqrt{23}$ $21.79 \pm ( 2.5) ( 8.22)/$ $\displaystyle \sqrt{23}$ $21.79 \pm ( 2.492) ( 8.22)/$ $\displaystyle 2\,\sqrt{6}$ $21.79 \pm ( 2.5) ( 8.22)/$ $\displaystyle 2\,\sqrt{6}$

  2. An experimenter is interested in the hypothesis testing problem

    $\displaystyle H_0: \: \mu = 0.75$    versus $\displaystyle H_A: \: \mu > 0.75 $

    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, and that the sample mean $\bar{X}$ = 0.79 and the sample standard deviation is $S$ = 0.376. Use the significance level $\alpha = 0.01$, and find the correct statement.

    Since $T = ( 0.79 - 0.75)/$ (0.094) = 0.426 is less than 2.947, the null hypothesis cannot be rejected. Since $T = ( 0.79 - 0.75)/$ ( $\displaystyle {{0.376}\over{\sqrt{15}}}$) = 0.412 is less than 2.602, the null hypothesis can be rejected. Since $T = ( 0.79 - 0.75)/$ (0.094) = 0.426 is less than 2.947, the null hypothesis can be rejected. Since $T = ( 0.79 - 0.75)/$ (0.094) = 0.426 is less than 2.602, the null hypothesis cannot be rejected. Since $T = ( 0.79 - 0.75)/$ (0.094) = 0.426 is less than 2.602, the null hypothesis can be rejected. Since $T = ( 0.79 - 0.75)/$ ( $\displaystyle {{0.376}\over{\sqrt{15}}}$) = 0.412 is less than 2.602, the null hypothesis cannot be rejected.

  3. A consumer agency suspects that a pet food company may be underfilling packages for one of its brands. The package label states “ 1000 grams net weight,” and the president of the company claims the average weight is greater than the amount stated. For a random sample of 16 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 942.302 grams and the sample standard deviation is $S$ = 51.77. Use the significance level $\alpha = 0.01$, and find the correct statement.

    $T = ( 942.302 - 1000)/$ ( $\displaystyle {{51.77}\over{\sqrt{15}}}$) = -4.316 is less than -2.602, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1000$. Thus, the evidence for underfilling is statistically not significant . $T = ( 942.302 - 1000)/$ (12.9425) = -4.458 is less than -2.602, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1000$. Thus, the evidence for underfilling is statistically significant . $T = ( 942.302 - 1000)/$ ( $\displaystyle {{51.77}\over{\sqrt{15}}}$) = -4.316 is less than -2.602, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 1000$. Thus, the evidence for underfilling is statistically significant . $T = ( 942.302 - 1000)/$ (12.9425) = -4.458 is less than -2.602, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 1000$. Thus, the evidence for underfilling is statistically not significant .

  4. 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 23 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 $\vert T\vert < 1.717$ The null hypothesis is rejected when $\vert T\vert > 1.717$ The null hypothesis is rejected when $\vert T\vert > 1.714$ The null hypothesis is rejected when $\vert T\vert < 1.714$ The null hypothesis is rejected when $\vert T\vert > 1.321$ The null hypothesis is rejected when $\vert T\vert < 1.321$

  5. 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 10 wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 213.1 and the sample standard deviation is $S$ = 96.14. Use the significance level $\alpha = 0.05$, and find the correct statement.

    Since $\vert T\vert = \vert 213.1 - 200\vert/$ ( $\displaystyle {{96.14}\over{\sqrt{10}}}$) = 0.431 is less than 1.833, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 213.1 - 200\vert/$ ( $\displaystyle {{96.14}\over{\sqrt{10}}}$) = 0.431 is less than 2.262, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 213.1 - 200\vert/$ (32.04666666666667) = 0.409 is less than 2.262, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 213.1 - 200\vert/$ ( $\displaystyle {{96.14}\over{\sqrt{10}}}$) = 0.431 is less than 2.262, the null hypothesis can be rejected. Since $\vert T\vert = \vert 213.1 - 200\vert/$ (32.04666666666667) = 0.409 is less than 2.262, the null hypothesis can be rejected. Since $\vert T\vert = \vert 213.1 - 200\vert/$ ( $\displaystyle {{96.14}\over{\sqrt{10}}}$) = 0.431 is less than 1.833, the null hypothesis can be rejected.


Department of Mathematics
Last modified: 2026-05-20