document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169146>Question 1169146</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Workers at a certain soda drink factory collected data on the volumes​ (in ounces) of a simple random sample of 19 cans of the soda drink. those volumes have a mean of 12.19 and a standard deviation of 0.08 oz and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.95 oz and 12.55 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.15 oz.  Use the sample data to test the claim that the population of volumes has a standard deviation less than  0.15 oz . Use a 0.05 significance level. Complete parts​ (a) through​ (d) below.<BR>\n" );
document.write( "ho: o=0.15 oz<BR>\n" );
document.write( "h1:o < 0.15 oz<BR>\n" );
document.write( "Need to find x to the second power<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851399 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1980)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851399\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this hypothesis test for the standard deviation:\r<BR>\n" );
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document.write( "**1. Define the Hypotheses and Significance Level:**\r<BR>\n" );
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document.write( "* Null Hypothesis (H₀): σ = 0.15 oz (The population standard deviation is 0.15 oz)<BR>\n" );
document.write( "* Alternative Hypothesis (H₁): σ < 0.15 oz (The population standard deviation is less than 0.15 oz)<BR>\n" );
document.write( "* Significance level (α) = 0.05\r<BR>\n" );
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document.write( "**2. Gather the Data:**\r<BR>\n" );
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document.write( "* Sample size (n) = 19<BR>\n" );
document.write( "* Sample mean (x̄) = 12.19 oz (This is actually irrelevant for the chi-square test for standard deviation)<BR>\n" );
document.write( "* Sample standard deviation (s) = 0.08 oz<BR>\n" );
document.write( "* Population standard deviation (σ₀) = 0.15 oz (from the null hypothesis)\r<BR>\n" );
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document.write( "**3. Calculate the Chi-Square Test Statistic (χ²):**\r<BR>\n" );
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document.write( "* The formula for the chi-square test statistic is:\r<BR>\n" );
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document.write( "    χ² = (n - 1) * s² / σ₀²\r<BR>\n" );
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document.write( "* Plug in the values:\r<BR>\n" );
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document.write( "    χ² = (19 - 1) * (0.08)² / (0.15)²<BR>\n" );
document.write( "    χ² = 18 * 0.0064 / 0.0225<BR>\n" );
document.write( "    χ² = 18 * (64/225)<BR>\n" );
document.write( "    χ² = 18 * 0.284444...<BR>\n" );
document.write( "    χ² = 5.12\r<BR>\n" );
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document.write( "**4. Determine the Degrees of Freedom:**\r<BR>\n" );
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document.write( "* Degrees of freedom (df) = n - 1 = 19 - 1 = 18\r<BR>\n" );
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document.write( "**5. Find the Critical Chi-Square Value:**\r<BR>\n" );
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document.write( "* Since this is a left-tailed test (H₁: σ < 0.15), we need to find the critical chi-square value from a chi-square distribution table with 18 degrees of freedom and an alpha level of 0.05.<BR>\n" );
document.write( "* Using a chi-square table or calculator, the critical chi-square value is approximately 9.390.\r<BR>\n" );
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document.write( "**6. Compare the Test Statistic to the Critical Value:**\r<BR>\n" );
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document.write( "* Our calculated chi-square test statistic (5.12) is less than the critical chi-square value (9.390).\r<BR>\n" );
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document.write( "**7. Make a Decision:**\r<BR>\n" );
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document.write( "* Because our calculated chi-square value (5.12) falls within the critical region (it is less than the critical value), we reject the null hypothesis.\r<BR>\n" );
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document.write( "**8. Conclusion:**\r<BR>\n" );
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document.write( "* There is significant evidence at the α = 0.05 level to conclude that the population standard deviation of the soda drink volumes is less than 0.15 oz.\r<BR>\n" );
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document.write( "**Therefore, x to the second power is 5.12**<BR>\n" );
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