A software company develops and markets a popular business simulation/modeling program. A random number generator contained in the program provides random values from various probability...


A software company develops and markets a popular business simulation/modeling program. A random number generator contained<br>in the program provides random values from various probability distributions. The software design group would like to validate that the<br>program is properly generating random numbers. Accordingly, they generated 5,000 random numbers from a normal distribution and<br>grouped the results into the accompanying frequency distribution shown below. The sample mean and sample standard deviation are<br>90 and 10, respectively. (You may find it useful to reference the appropriate table: chi-square table or F table)<br>Value<br>Frequency<br>Under 60<br>8.<br>60 up to 70<br>70 up to 80<br>80 up to 90<br>90 up to 100<br>109<br>682<br>1,757<br>1,694<br>100 up to 110<br>110 up to 120<br>649<br>94<br>120 or more<br>Total = 5,000<br>a. Using the goodness-of-fit test for normality, state the competing hypotheses to test if the random numbers generated do not follow<br>the normal distribution.<br>O Ho: Random numbers are normally distributed with a mean of 90 and a standard deviation of 1O. HA: Random numbers are not<br>normally distributed with a mean of 90 and a standard deviation of 10.<br>O Ho: Random numbers are not normally distributed with a mean of 90 and a standard deviation of 10. HA: Random numbers are<br>normally distributed with a mean of 90 and a standard deviation of 10.<br>IN<br>< Prev<br>5 of 10<br>Next ><br>here to search<br>(?<br>

Extracted text: A software company develops and markets a popular business simulation/modeling program. A random number generator contained in the program provides random values from various probability distributions. The software design group would like to validate that the program is properly generating random numbers. Accordingly, they generated 5,000 random numbers from a normal distribution and grouped the results into the accompanying frequency distribution shown below. The sample mean and sample standard deviation are 90 and 10, respectively. (You may find it useful to reference the appropriate table: chi-square table or F table) Value Frequency Under 60 8. 60 up to 70 70 up to 80 80 up to 90 90 up to 100 109 682 1,757 1,694 100 up to 110 110 up to 120 649 94 120 or more Total = 5,000 a. Using the goodness-of-fit test for normality, state the competing hypotheses to test if the random numbers generated do not follow the normal distribution. O Ho: Random numbers are normally distributed with a mean of 90 and a standard deviation of 1O. HA: Random numbers are not normally distributed with a mean of 90 and a standard deviation of 10. O Ho: Random numbers are not normally distributed with a mean of 90 and a standard deviation of 10. HA: Random numbers are normally distributed with a mean of 90 and a standard deviation of 10. IN < prev="" 5="" of="" 10="" next=""> here to search (?
b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3<br>decimal places.)<br>Test statistic<br>b-2. Fine the p-value?<br>O p-value <0.01<br>O 0.01B p-value <0.025<br>O 0.025 B pvalue< 0.05<br>O 0.05 3 p-value <0.10<br>O p-value 2o.10<br>c-1. What is the conclusion to the test?<br>< Prev.<br>5 of 10<br>Next ><br>e to search<br>

Extracted text: b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) Test statistic b-2. Fine the p-value? O p-value <0.01 o="" 0.01b="" p-value=""><0.025 o="" 0.025="" b="">< 0.05="" o="" 0.05="" 3="" p-value=""><0.10 o="" p-value="" 2o.10="" c-1.="" what="" is="" the="" conclusion="" to="" the="" test?="">< prev.="" 5="" of="" 10="" next=""> e to search
Jun 07, 2022
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