Part 1 of 4 A study is done to determine which of two soft drinks has more sugar. There are 25 cans of Beverage A in the researchers' sample and 15 cans of Beverage B. The mean amount of sugar in Beverage A is 34.7 grams per can with a sample standard deviation of 0.7 grams. The mean amount of sugar in Beverage B is 33.9 grams with a standard deviation of 0.6 grams. The researchers believe that Beverage A has less sugar than Beverage B, on average. Both populations are normally distributed. Conduct a hypothesis test using a 2% level of significance. Step 1: State the null and alternative hypotheses. Ho: HA - HB 2 Ha: HA - HB (So we will be performing a left-tailed test.) Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of the differences of sample means. with distribution mean The differences of sample means are t-distributed and distribution standard deviation Question Help: D Post to forum

Part 1 of 4 A study is done to determine which of two soft drinks has more sugar. There are 25 cans of Beverage A in the researchers' sample and 15 cans of Beverage B. The mean amount of sugar in Beverage A is 34.7 grams per can with a sample standard deviation of 0.7 grams. The mean amount of sugar in Beverage B is 33.9 grams with a standard deviation of 0.6 grams. The researchers believe that Beverage A has less sugar than Beverage B, on average. Both populations are normally distributed. Conduct a hypothesis test using a 2% level of significance. Step 1: State the null and alternative hypotheses. Ho: HA - HB 2 Ha: HA - HB (So we will be performing a left-tailed test.) Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of the differences of sample means. with distribution mean The differences of sample means are t-distributed and distribution standard deviation Question Help: D Post to forum

Name: Part 1 of 4A study is done to determine which of two soft drinks has
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Part 1 of 4A study is done to determine which of two soft drinks has more sugar. There are 25 cans ofBeverage A in the researchers' sample and 15 cans of Beverage B. The mean amount of sugar inBeverage A is 34.7 grams per can with a sample standard

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Forty refrigerators from two different models of a refrigerator manufacturer are compared to see if the smaller model uses less energy than the larger model. The energy was measured in kilowatt hours per month (kwh/mo). From years of testing these models, it is known that the standard deviation of kwh/mo is 34 kwh/mo for the smaller model and 40 kwh/mo for the larger one. From this survey, the sample means and standard deviations are 125 and 34, and 90 and 40, respectively. If we conduct a hypothesis test, the p-value is Group of answer choices 0.0000 0.0010 0.0004 0.0002
A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in the sample for Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in the sample Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. What is the p-value? Round to 4 decimal places.
For a study conducted by the research department of a pharmaceutical company, 235 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.80 with a standard deviation of $4.30 . A random sample of 250 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.20 with a standard deviation of $4.00 . As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 90% confidence interval for the difference −μ1μ2 between the mean spending on prescription allergy relief medication ( μ1 ) and the mean…
Are low-fat diets or low-carb diets more effective for weight loss? A sample of 61 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 2.1 kilograms with a sample standard deviation of 4.71 kilograms. A second sample of 63 subjects went on a low-fat diet. Their sample mean weight loss was 3.7 kilograms with a standard deviation of 5.67 kilograms. Can you conclude that the mean weight loss of subjects having low-carb diets is less than the mean weight loss of subjects having low-fat diets? Let μ1 denote the mean weight loss on the low-carb diet and μ2 denote the mean weight loss on the low-fat diet. Use the =α0.05 level and the critical value method. 1) State the appropriate null and alternate hypotheses. 2)Choose a significance level and find the critical value. 3)Compute the test statistic. To compute the test statistic, we use the formula 4)Determine whether to reject H0.
A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 38 participants in the treatment group lowered their cholesterol levels by a mean of 23.2 points with a standard deviation of 4.4 points. The 44 participants in the control group lowered their cholesterol levels by a mean of 21.5 points with a standard deviation of 2.9 points. Assume that the population variances are not equal and test the company's claim at the 0.01 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
I need help, I keep getting it wrong and only have one more try for each.
A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo.In a random sample of 49 floors of WaxWin and 35 of WaxCo. WaxWin had a mean lifetime of 25.3 and WaxCo had a mean lifetime of 27.3.The population standard deviation for WaxWin is assumed to be 7.8 and the population standard deviation for WaxCo is assumed to be 10.2.Perform a hypothesis test using a significance level of 0.05 to help him decide.Let WaxWin be sample 1 and WaxCo be sample 2.
I need help understanding how to solve the question with a black box around it.
Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is μ=20 inches. However, a survey reported that of a simple random sample of 37 fish caught, the mean length was 17.5 inches, with an estimated standard deviation of 4.3 inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than μ=20 inches? Use α=0.05. What is the value of the test statistic? (Round your answer to two decimal places.)Find the P-value. (Round your answer to four decimal places.)Based on your results, will you reject or fail to reject the null hypothesis? Are the data statistically significant at level αα? A) At the α=0.05α=0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. B) At the α=0.05α=0.05 level, we reject the null hypothesis and conclude the data are statistically significant. C)…
3. At a land farm, a farmer desired to test the effect of a given fertilizer on soybean production. He chose 24 plots of land of which are equal surface area. He treated 12 plots with the fertilizer and the others were untreated. Otherwise the conditions were the same. The treated plots produced soybean with mean yield 5.1 bushels and a standard deviation of 0.36 bushels, while the untreated plots had mean yield 4.8 bushels and a standard deviation of 0.40 bushels. (a) Can we conclude that there is a significant improvement in soybean production because of the fertilizer if a significance level of 1% is used? (b) What is the P-value of the test?
Average adult Americans are about one inch taller but nearly a whopping 25 pounds heavier than they were in 1960, according to a report from the Centers for Disease Control and Prevention (CDC). City A is considered one of America's healthiest cities. Is the weight gain since 1960 similar in city A? A sample of n=25 adults suggested a mean increase of 22 pounds with a standard deviation of 9.4 pounds. Is City A significantly different in terms of weight gain since 1960? Complete and interpret 5 step hypothesis testing by hand at a 5% level of significance using a Z-test. Show each step and your work.
Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women