5 ppl sample Mean 22.2 A deviation 5.8 Confidence 90%
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- see attachmentPlease explain all subpartsOInstructor-created question Question Help ▼ In a random sample of n = 50 patients, the mean creatine level .x = 1.2 mg/dL with unreported sample standard deviation. The mean creatine level u in the population of all patients is unknown with standard deviation a = 0.5 mg/dL. With 95% confidence compare the mean creatine level of the patient population to the 1.O mg/dL mean creatine level of the general population using the p-value method: Test Ho: u= 1.0 mg/dL Patient creatine level is same as that of general population. Versus Ha: u#1.0 mg/dL Patient creatine level is different than that of general population. Use all digits showing on your calculator in your intermediate calculations but show only the requested digits to the right of the decimal point in your final answers. (a) test statistic showing 2 digits to the right of the decimal: (b) p-value showing 4 digits to the right of the decimal: (c)Conclusion: O A. Reject Ho with 95% confidence and conclude that the mean creatine…
- When calculating confidence intervals for differences between two population means, the sample size does matter. True Falseknown standard deviationAssum a sample is used to estimate a populatation mean. Find the 95% confidence interval for a sample size of 41 with a mean of 64.2 and a standard deviation of 13.4. Enter answer as an open interval (i.e. parentheses) accurate to 3 decimal places.
- Verdadero=true falso=falseWe want to estimate the true mean number of ounces of water a high school athlete drinks each day at a 95% confidence level. A random sample of 15 athletes from a school is selected and their typical daily water consumption is reported below: 64, 72, 80, 88, 128, 128, 128, 132, 145, 155, 160, 175, 180, 200, 225 What is the value of the standard error of the mean? (When calculating, use the standard deviation of the sample rounded to 3 decimal places and then round your standard error answer to 3 decimal places.) =best point estimate for 0.678>p>0.788 with a 95% confidence interval
- If a sample of 25 fish from my pond has mean weight 157 grams with a standard deviation of 8.3 grams and we know the population standard deviation for the fish in my pond is actually 9.4 grams, then WHAT IS the MARGIN OF ERROR in the 99 percent confidence interval for true mean weight of fish in my pond? Group of answer choicesWomen N Mean Standard Deviation New Drug 40 38.88 3.97 Placebo 41 39.24 4.21 Men N Mean Standard Deviation New Drug 10 45.25 1.89 Placebo 9 39.06 2.22 There is no statistical difference between the new drug and the placebo in women. Is there a significant difference between the placebo and the new drug among men in this study? The pooled standard deviation is equal to 2.05. Please complete the 5 steps of hypothesis testing. Remember that the n of both groups is less than 30.Determine Range sample standard deviation sample variance what does the results tell us?
