<p>Unit 4 Application Assignment Hypothesis Tests</p><p>Instructions Please read the following scenario and complete the requirements below. Use the data set below and the results you developed from your previous Application Assignments to complete this assignment. You may create your assignment using either Microsoft Word or Excel. When you have completed your assignment, use the following naming convention to name your file: The course number, your first initial, your last name, the unit number and the assignment. For example: MTH140_C_Smith_U4 App Assign.doc. Scenario </p><p>Recall the information from Candy Count Assignment. </p><p>Using that information, write a pair of hypothesis for each of the claims and perform the hypothesis test. Be sure to list α, critical values, standardized test statistics, P-values, and decision. Interpret the results. Assume standard deviations are known and the candies are normally distributed. </p><p>•With 90% confidence, the average number of red candies is 3.</p><p>Answer:</p><p>Null hypothesis The average number of red candies is not significantly different from the population mean of 3 or</p><p>Ho:  = 3 </p><p>Alternative hypothesis The average number of red candies is significantly different from the population mean of 3 or </p><p>Ha:  ≠ 3 </p><p>Using One sample Z test (since the population standard deviation is known), and alpha = 0.10, (two-tailed test)</p><p>The critical z values are z =  1.6449</p><p>Test Statistic:</p><p> z = (x – ) / (s / n)</p><p>= (2.6 – 3) / (2.27 / 10)</p><p>= – 0.5572</p><p>The Computed z value is – 0.5572 and p – value = 0.5774 Decision:</p><p>Since the computed z value (– 0.5572) is within the 2 critical z values ( 1.6449), then there is no sufficient evidence to reject the null hypothesis.</p><p>Conclusion/Result The average number of red candies is not significantly different from the population mean of 3 or</p><p>Ho:  = 3 at alpha = 0.10 (2 tailed). </p><p>The average number of red candies is 3.</p><p>•With 99% confidence, the average number of blue candies is less than 2.5. </p><p>Answer:</p><p>Null hypothesis</p><p>The average number of blue candies is not significantly less than 2.5 or Ho:  = 2.5 </p><p>Alternative hypothesis</p><p>The average number of blue candies is significantly less than 2.5 or Ha:  < 2.5 </p><p>Using One sample Z test (since the population standard deviation is known), and alpha = 0.01, (left-tailed test)</p><p>The critical z value is z = – 2.3263</p><p>Test Statistic:</p><p> z = (x – ) / (s / n)</p><p>= (3.6 – 2.5) / (1.78 / 10)</p><p>= 1.9542</p><p>The Computed z value is 1.9542 and p – value = 0.9747</p><p>Decision:</p><p>Since the computed z value (1.9542) is not less than the critical z value (–2.3263), then there is no sufficient evidence to reject the null hypothesis.</p><p>Conclusion/Result</p><p>The average number of blue candies is not significantly less than 2.5 or Ho:  = 2.5 at alpha = 0.01 (left-tailed). </p><p>The average number of blue candies is not less than 2.5 •With 95% confidence, the variance of the number of green candies is greater than or equal to 2.5. </p><p>Answer:</p><p>Null hypothesis The variance of the number of green candies is not significantly less than 2.5 or Ho: 2 ≥ 2.5 </p><p>Alternative hypothesis The variance of the number of green candies is significantly less than 2.5 or Ha: 2 < 2.5 </p><p>Using Chi Square test for one sample variance, and alpha = 0.05, (left-tailed test)</p><p>The critical 2 value at df = n – 1 = 10 – 1 = 9 is 2 = 3.3251</p><p>Test Statistic:</p><p>2 = (n – 1)s2 / 2</p><p>= (10 – 1)*1.2889 / 2.5</p><p>= 4.64</p><p>The Computed 2 value is 4.64 and p – value = 0.1355</p><p>Decision:</p><p>Since the computed 2 value (4.64) is not less than the critical 2 value (3.3251), then there is no sufficient evidence to reject the null hypothesis.</p><p>Conclusion/Result The variance of the number of green candies is not significantly less than 2.5 or Ho: 2 ≥ 2.5 at alpha = 0.05 (left-tailed). </p><p>The variance of the number of green candies is greater than or equal to 2.5. </p><p>•With 90% confidence, the variance of the number of red candies is 3. Attached is the Candy Count Answer:</p><p>Null hypothesis The variance of the number of red candies is not significantly different from 3 or Ho: 2 = 3 </p><p>Alternative hypothesis The variance of the number of red candies is significantly different from 3 or Ha: 2 ≠ 2.5 </p><p>Using Chi Square test for one sample variance, and alpha = 0.10, (2-tailed test)</p><p>The critical 2 values at df = n – 1 = 10 – 1 = 9 are 2 = 3.3251 and 2 = 16.9190</p><p>Test Statistic:</p><p>2 = (n – 1)s2 / 2</p><p>= (10 – 1)*5.1556 / 3</p><p>= 15. 4668</p><p>The Computed 2 value is 15. 4668 and p – value = 0.1578</p><p>Decision:</p><p>Since the computed 2 value (15. 4668) is within the 2 critical 2 values (3.3251, 16.9190), then there is no sufficient evidence to reject the null hypothesis.</p><p>Conclusion/Result The variance of the number of red candies is not significantly different from 3 or Ho: 2 = 3 at alpha = 0.10 (2-tailed). </p><p>The variance of the number of red candies is 3. </p>