A herbal extract is claimed to lower the blood glucose level. 10 prediabetic subjects were give capsules containing lactose for1 month. Another sample of 10 subjects were given capsules containing dried herbal extract for one month. Assuming the samples were taken from populations of known equal standard deviations of 10 and the question of interest was does the herbal extract result in significant BGL reductio compared to placebo. Based on HO: Pherbal- Hplacebo20 at sig. level of 0.1, the null hypothesis states that

A herbal extract is claimed to lower the blood glucose level. 10 prediabetic subjects were give capsules containing lactose for1 month. Another sample of 10 subjects were given capsules containing dried herbal extract for one month. Assuming the samples were taken from populations of known equal standard deviations of 10 and the question of interest was does the herbal extract result in significant BGL reductio compared to placebo. Based on HO: Pherbal- Hplacebo20 at sig. level of 0.1, the null hypothesis states that

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Description: A herbal extract is claimed to lower the blood glucose level. 10 prediabetic subjectswere give capsules containing lactose for 1 month. Another sample of 10 subjectswere given capsules containing dried herbal extract for one month. Assuming thesampl

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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An article reports an investigation into the toxicity of bilirubin on several cell lines. Ten sets of human liver cells and 5 sets of mouse fibroblast cells were placed into solutions of bilirubin in albumin with a 1.4 bilirubin/albumin molar ratio for 24 hours. In the 10 sets of human liver cells, the average percentage of cells surviving was 53.9 with a standard deviation of 10.7. In the 5 sets of mouse fibroblast cells, the average percentage of cells surviving was 73.1 with a standard deviation of 9.1. Let μXμX represent the population mean for the mouse cells and let μYμY represent the population mean for the human cells. Find a 98% confidence interval for the difference μX−μYμX−μY . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places. The 98% confidence interval is ( , ).
A random sample of people who tried 3 types of diet is collectedto determin if mean weight loss differs across diet types. Perform inference at significance level 0.05. If it is needed, perform inference taking age as a covariate. Weight Loss: (3.841, 2.5341, 8.6165, 7.3756, 5.4883, 10.591, 12.2113, 3.6752, 10.2109, 12.8081, 14.9185, 9.1484, 14.6684, 17.3383, 14.3372) Diet Type: (typе1, "tуpe1', 'type1, "type1', "type1', "tуpe2", "type2, tуpe2", 'type2", "type2, "type3', typе3", "typе3', typе3', 'type3") Age: (30, 30, 24, 30, 31, 30, 32, 30, 28, 36, 46, 50, 50, 48, 46) After adjusting for age, there is significant evidence that there are differences of mean weight loss across 3 diet types. There is not significant evidence that there are differences of mean weight loss across 3 diet types. After adjusting for age, there is not significant evidence that there are differences of mean weight loss across 3 diet types. We cannot determine whether there are differences of mean weight loss…
A company that produces top-of-the-line batteries claims that its batteries are good, on average, for more than 65 months. A consumer protection agency tested 25 such batteries to check this claim. It found that the mean life of these 25 batteries is 63.4 months, and the standard deviation is 3 months. If we test the company’s claim at a 5% significance level, what is the p-value? Assume normality. A company that produces top-of-the-line batteries claims that its batteries are good, on average, for more than 65 months. A consumer protection agency tested 25 such batteries to check this claim. It found that the mean life of these 25 batteries is 63.4 months, and the standard deviation is 3 months. If we test the company’s claim at a 5% significance level, what is the p-value? Assume normality. 0.0067 0.0135 0.9962 0.9933
In the ANOVA we are comparing more than just two means. When comparing just two means we set our alpha at .05. If we were comparing multiple means (more than two), two at a time (such as in a t-Test), we would have a greater likelihood of committing a Type l error. The probability of making a Type I error when comparing all means in an experiment is called: O compounded alpha adjustment (CAA) O standard error of the levels O multiple-means error rate O experiment-wise error rate
The manufacturer of the X-15 steel-belted redial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Crosset Truck Company bought a sample of 48 tires and found that the mean mileage for its trucks is 59,000 miles (i.e. less than the manufacturer’s claim) with a standard deviation of 5,000 miles. Test the manufacturer’s claim at a significance level (α) equals 5%. Would your answer in (1) change if significance level become 10%? Would your answer in (1) change if the probability of type I error equals zero? Justify your answer If the same sample results had been obtained from a random sample of 125 tires, could your decision about the manufacturer’s claim changed using α=0.01 and comment on the results. Based on the sample size in (4), construct a 95% confidence level for the mean of mileage and comment on the results.
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A health administration recommends that individuals consume 1000 mg of calcium daily. After an advertising campaign aimed at male teenagers, a dairy association states that male teenagers on average consume more than the recommended daily amount of calcium. To support this statement, the association obtained a random sample of 40 male teenagers and found that the mean amount of calcium consumed was 1020 mg, with a standard deviation of 100 mg. Is there significant evidence to support the statement of the association at the α=0.05 level of significance?
Recently, a simple random sample of 572 adult males showed that 124 of them smoke. It has been established that in 2008. 20.4% of adults smoke. Use a 0.05 significance level to test the claim that the rate of smoking by adult males is now the same as in 2008. What is the value of the critical z tabular value?
In a sample of 87 people who have had strokes, the average cholesterol level was 258 with a standard deviation of 41.6. In order to test the hypothesis (at the 10% level of significance) that the average cholesterol level of people who have had strokes was at least 240, what is the critical value?
In a study of microplastics in our environment, it was determined that there are 5.25 trillion plastic pieces floating in the ocean, weighing more than 250,000 tons. In a recent expedition to examine whether increases in plastic production have led to changes in ocean microplastics, 16 samples were taken at various locations in the Pacific Ocean. It was determined that the average concentration was 9.5 particles per liter of ocean water with a sample standard deviation of 6. In one of the earliest research expeditions to study microplastics in the ocean conducted in 2009, it was determined that the average concentration of microplastics was 6 particles per liter (consider this the population or reference mean). What is the critical value for the two-tailed alternative hypothesis at alpha = 5%? (Hint: This is the value in the table that you will compare your test statistic to in order to determine the outcome of the hypothesis test). 2.131 1.746 2.120 1.753
An airline company is interested in improving customer satisfaction rate from the 66% currently claimed. The company sponsored a survey of 262 customers and found that 184 customers were satisfied. What is the test statistic z? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is higher than the claim by the company at a significance level of α=0.01 (Y/N)