a goodness of fit test was run to determine if a sample of data was normally distributd. th theoretcal distributiones showed a posibility of 17 expecteed values. at the .05 level the approprotae table statistics is : 22.36, 26.29, 24.99, 23.89
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a goodness of fit test was run to determine if a sample of data was normally distributd. th theoretcal distributiones showed a posibility of 17 expecteed values. at the .05 level the approprotae table statistics is : 22.36, 26.29, 24.99, 23.89
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- Maple trees in a large forest have heights that are normally distributed with a mean of 95 feet and a standard deviation of 4.7 feet. One tree has a height of 92 feet. What is the Z-score of this height?Patients with two kinds of diagnoses were randomly assigned to one of three types of therapy and the effectiveness of the therapy was measure a 1-to-15 scale (with a higher number indicating greater effectiveness). There were two patients per cell. Use the data available below to comple parts (a) through (d) below. Click the icon to view the data table. Click here to view page 1 of the table of cutoff values for the F distribution. Click here to view page 2 of the table of cutoff values for the F distribution. Click here to view page 3 of the table of cutoff values for the F distribution. Click here to view page 4 of the table of cutoff values for the F distribution (a) Carry out the analysis of variance (use the 0.05 significance level). Complete the analysis of variance table for the data set below. SS MS F Source Columns Rows Interaction Within Total (Round to two decimal places as needed.) Data Table Diagnosis I Diagnosis II Therapy A 5 1 7 Print Therapy B 7 12 Done Therapy C 4282…A company selling a new type of laptop battery advertises that it has a méan lifetime of 5 years. If you were in charge of quality control at the factory, which one of the following standard deviations of battery life spans would you prefer? 2 months. O 2 years O 5 years 27
- Weights (kg) of poplar trees were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective? LOADING... Click the icon to view the data table of the poplar weights. No Treatment Fertilizer Irrigation Fertilizer and Irrigation 1.24 0.99 0.09 0.93 0.65 0.94 0.63 1.41 0.41 0.55 0.13 1.06 0.15 0.51 0.88 1.49 1.31 1.01 0.84 Determine the null and alternative hypotheses. H0: ▼ mu 1 greater than mu 2 greater than mu 3 greater than mu 4μ1>μ2>μ3>μ4 Exactly two of the population means are equal.Exactly two of the population means are equal. Not all of the population means are equal.Not all of the population means…Entrance exam scores at Cititon University are normally distributed with a mean of 552 and a standard deviation of 29. If scholarships are awarded to the top 10% of scores, what is the minimum score needed to guarantee a scholarship?An air condition company will spend an average of 64 minutes with a standard deviation of 19 minutes on a service call. Based on this, and assuming a normal distribution, 5% of service calls will take more than how much time to complete? Approximately 66.29 minutes O Approximately 32.75 minutes O Approximately 39.41 minutes O Approximately 95.25 minutes O Approximately 77.23 minutes
- form a confidence interval for mean, proportion, and standard deviation.A sample of underweight babies was fed a special diet and the following weight gains (lbs.) were observed at the end ofthree months.6.7 5.8 4.8 3.6 4.7 4.6 5.2 5.9 4.5The mean and standard deviation respectively are:a. 5.09, 0.92 b. 5.1, 0.87 c. 5.08, 8.7 d. 5.09, 0.87 e. none of these
