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Solve the Problem Using Technology, Perform the Following Hypothesis Test: at the 5

Question 46

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Solve the problem. Use the critical-value approach.
-A machine that fills soda bottles is supposed to fill them to a mean volume of 16.2 fluid ounces. A random sample of 20 filled bottles produced the following volumes in fluid ounces: 16.315.916.715.317.116.43.915.916.23.416.48.215.516.516.016.315.816.716.515.5 These data are summarized on the following histogram: \begin{array}{l}\begin{array} { l l l l l l l l l l } 16.3 & 15.9 & 16.7 & 15.3 & 17.1 & 16.4 & 3.9 & 15.9 & 16.2 & 3.4 \\16.4 & 8.2 & 15.5 & 16.5 & 16.0 & 16.3 & 15.8 & 16.7 & 16.5 & 15.5\end{array}\\\text { These data are summarized on the following histogram: }\end{array}
 Solve the problem. Use the critical-value approach. -A machine that fills soda bottles is supposed to fill them to a mean volume of 16.2 fluid ounces. A random sample of 20 filled bottles produced the following volumes in fluid ounces:  \begin{array}{l} \begin{array} { l l l l l l l l l l }  16.3 & 15.9 & 16.7 & 15.3 & 17.1 & 16.4 & 3.9 & 15.9 & 16.2 & 3.4 \\ 16.4 & 8.2 & 15.5 & 16.5 & 16.0 & 16.3 & 15.8 & 16.7 & 16.5 & 15.5 \end{array}\\ \text { These data are summarized on the following histogram: } \end{array}     Using technology, perform the following hypothesis test: at the 5% significance level,determine Whether the fill volume is less than the supposed value. Comment on the appropriateness of the Test.  A)  Test statistic:  t = - 1.8063 ; Critical value:  - 2.0921 . Since the test statistic is greater than the critical value, do not reject the null hypothesis  \mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } . There is insufficient evidence to conclude that the fill volume is below  16.2  oz. The conclusion is on sound statistical ground. B)  Test statistic:  t = - 1.8063 ; Critical value:  - 1.7254 . Since the test statistic is less than the critical value, do not reject the null hypothesis  \mathrm { H } _ { 0 } : \mu =   16.2 \mathrm { oz } . There is insufficient evidence to conclude that the fill volume is below  16.2  oz. However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion. C)  Test statistic:  t = - 1.8063 ; Critical value;  - 2.0921 . Since the test statistic is greater than the critical value, do not reject the null hypothesis  \mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } . There is insufficient evidence to conclude that the fill volume is below  16.2  oz. However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion. D)  Test statistic:  t = - 1.8063 ; Critical value:  - 1.729 . Since the test statistic is less than the critical value, reject the null hypothesis  \mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } . There is sufficient evidence to conclude that the fill volume is below  16.2 \mathrm { oz } . However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion.
Using technology, perform the following hypothesis test: at the 5% significance level,determine Whether the fill volume is less than the supposed value. Comment on the appropriateness of the Test.


A) Test statistic: t=−1.8063t = - 1.8063 ; Critical value: −2.0921- 2.0921 .
Since the test statistic is greater than the critical value, do not reject the null hypothesis H0:μ=16.2oz\mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } . There is insufficient evidence to conclude that the fill volume is below 16.216.2 oz. The conclusion is on sound statistical ground.
B) Test statistic: t=−1.8063t = - 1.8063 ; Critical value: −1.7254- 1.7254 .
Since the test statistic is less than the critical value, do not reject the null hypothesis H0:μ=\mathrm { H } _ { 0 } : \mu = 16.2oz16.2 \mathrm { oz } .
There is insufficient evidence to conclude that the fill volume is below 16.216.2 oz. However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion.
C) Test statistic: t=−1.8063t = - 1.8063 ; Critical value; −2.0921- 2.0921 .
Since the test statistic is greater than the critical value, do not reject the null hypothesis H0:μ=16.2oz\mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } . There is insufficient evidence to conclude that the fill volume is below 16.216.2
oz. However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion.
D) Test statistic: t=−1.8063t = - 1.8063 ; Critical value: −1.729- 1.729 .
Since the test statistic is less than the critical value, reject the null hypothesis H0:μ=16.2oz\mathrm { H } _ { 0 } : \mu = 16.2 \mathrm { oz } .
There is sufficient evidence to conclude that the fill volume is below 16.2oz16.2 \mathrm { oz } . However, the data exhibits 3 outliers. Elimination of these outliers may alter the conclusion.

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