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 29 packages collected by the agency, the sample mean of the weights is $\bar{X}$ = 803.347 grams and the sample standard deviation is = 60.969. Use the significance level $\alpha = 0.01$ , and find the correct statement.
( $\displaystyle {{60.969}\over{\sqrt{29}}}$ ) = 0.296 is greater than -2.467, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 800$ . Thus, the evidence for underfilling is statistically not significant . ( $\displaystyle {{30.4845}\over{\sqrt{7}}}$ ) = 0.29 is greater than -2.467, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 800$ . Thus, the evidence for underfilling is statistically significant . ( $\displaystyle {{60.969}\over{\sqrt{29}}}$ ) = 0.296 is greater than -2.467, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu < 800$ . Thus, the evidence for underfilling is statistically significant . ( $\displaystyle {{30.4845}\over{\sqrt{7}}}$ ) = 0.29 is greater than -2.467, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu < 800$ . Thus, the evidence for underfilling is statistically not significant .
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 = 16 bottles of the chemical solution is obtained and their densities are measured, and that the sample mean $\bar{X}$ = 0.939 and the sample standard deviation is = 0.53. Use the significance level $\alpha = 0.1$ , and find the correct statement.
Since (0.1325) = 1.426 is less than 1.753, the null hypothesis can be rejected. Since ( $\displaystyle {{0.53}\over{\sqrt{15}}}$ ) = 1.381 is greater than 1.341, the null hypothesis can be rejected. Since ( $\displaystyle {{0.53}\over{\sqrt{15}}}$ ) = 1.381 is greater than 1.341, the null hypothesis cannot be rejected. Since (0.1325) = 1.426 is greater than 1.341, the null hypothesis cannot be rejected. Since (0.1325) = 1.426 is less than 1.753, the null hypothesis cannot be rejected. Since (0.1325) = 1.426 is greater than 1.341, the null hypothesis can be rejected.
A machine is set to cut metal plates to a length of 60 mm. The length of a random sample of 22 metal plates have a sample mean of $\bar{X}$ = 59.95 mm and a sample standard deviation of = 0.13 mm. Is there any evidence that the machine is miscalibrated? Use the significance level $\alpha = 0.01$ , and find the correct statement.
$\vert T\vert = \vert 59.95 - 60\vert/$ ( $\displaystyle {{0.13}\over{\sqrt{22}}}$ ) = 1.804 is less than 2.518, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 60$ . Thus, the evidence of miscalibration is statistically not significant . $\vert T\vert = \vert 59.95 - 60\vert/$ ( $\displaystyle {{0.13}\over{\sqrt{22}}}$ ) = 1.804 is less than 2.831, the null hypothesis can be rejected in favor of the alternative hypothesis $\mu \neq 60$ . Thus, the evidence of miscalibration is statistically significant . $\vert T\vert = \vert 59.95 - 60\vert/$ ( $\displaystyle {{0.13}\over{\sqrt{22}}}$ ) = 1.804 is less than 2.831, the null hypothesis cannot be rejected in favor of the alternative hypothesis $\mu \neq 60$ . Thus, the evidence of miscalibration is statistically not significant .
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 22 wool fiber bundles is obtained and their breaking strengths are measured. Suppose that the sample mean $\bar{X}$ = 436.36 and the sample standard deviation is = 63.72. Use the significance level $\alpha = 0.1$ , and find the correct statement.
Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{22}}}$ ) = 2.676 is greater than 1.323, the null hypothesis can be rejected. Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{22}}}$ ) = 2.676 is greater than 1.323, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{21}}}$ ) = 2.615 is greater than 1.721, the null hypothesis can be rejected. Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{21}}}$ ) = 2.615 is greater than 1.721, the null hypothesis cannot be rejected. Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{22}}}$ ) = 2.676 is greater than 1.721, the null hypothesis can be rejected. Since $\vert T\vert = \vert 436.36 - 400\vert/$ ( $\displaystyle {{63.72}\over{\sqrt{22}}}$ ) = 2.676 is greater than 1.721, the null hypothesis cannot be rejected.
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 = 22 bottles of the chemical solution is obtained and their densities are measured. For what values of the -statistic does the experimenter reject the null hypothesis with significance level $\alpha = 0.1$ ?
The null hypothesis is rejected when The null hypothesis is rejected when The null hypothesis is rejected when The null hypothesis is rejected when The null hypothesis is rejected when The null hypothesis is rejected when

Department of Mathematics
Last modified: 2026-09-06