A simple random sample of 101 course evaluations had a mean x¯x¯ = 3.90 with a standard deviation s = 0.52. Use a 0.05 significance level to test the claim that the population of course evaluations has a mean less than 4.00.
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw…
A: Here Given Sample Size=n= 36 Sample mean=0.3=x¯ Sample SD=s=1.84
Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw…
A: We have to identify correct pair of hypothesis.
Q: ..... What are the null and alternative hypotheses? OA. Ho: H= 0 mg/dL O B. Ho: u>0 mg/dL H1: µ0…
A: Given data : sample size, n =81 sample mean, x̄ = 0.8 sample standard deviation,s=…
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Q: garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus…
A: From the provided information, Sample size (n) = 81 Sample mean (x̅) = 0.9 Sample standard deviation…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects Were treated with raw…
A: Given,sample size(n)=81sample mean(x¯)=0.3standard deviation(S)=20.9α=0.10
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Q: A medical investigation claims that the average number of infections per week at a hospital in Town…
A: Solution-: Given: x¯=17.7,μ0=16.3,n=40,s=1.8,α=0.05 Claim: The average number of infections per week…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw…
A: Given information: Number of subjects used for test of the effectiveness of garlic n=36 The mean of…
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A: mu = 13xbar = 12.2s = 1.8n = 32
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with…
A: State the hypotheses. That is, there is no evidence that with the garlic treatment, the mean change…
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A simple random sample of 101 course evaluations had a
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- In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.1 and a standard deviation of 23.2. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? O A. Ho p=0 mg/dL O B. H, p=0mg/dL H, p0 mg/dL OC. Ho H>0 mg/dL O D. H, p=0 mg/dL. H, pz0 mg/dL. H, p<0 mg/dLAccording to a city's estimate, people on average arrive 1.5 hours early for domestic flights. The population standard deviation is known to be 0.5 hours. A researcher wanted to check if this is true so he took a random sample of 50 people taking domestic flights and found the mean time to be 2.0 hours early. At the 1% significance level, can you conclude that the amount of time people are early for domestic flights is more than what the city claims. Will you reject or not reject the null hypothesis and what is your conclusion in words? O Reject the null hypothesis; the city's claim is true. O Do not reject the null hypothesis; the city's claim is false and people arrive earlier than the city claims. O Reject the null hypothesis; the city's claim is false and people arrive earlier than the city claims. O Do not reject the null hypothesis; the city's claim is true.In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 2.33 . Use a 0.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0 . What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Upper H 0 : mu greater than0 mg/dL Upper H 1 : mu less than0 mg/dL B. Upper H 0 : mu equals0 mg/dL Upper H 1 : mu less than0 mg/dL C. Upper H 0 : mu…
- In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 20.8. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? O B. H,: H=0 mg/dL Ο Α. Η μ=0 mgldL H1: µ0 mg/dL O D. H,: µ> 0 mg/dL H: µ<0 mg/dL Determine the test statistic. (Round to two decimal places as needed.)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 womenA public bus company official claims that the mean waiting time for Bus # 14 during peak hours is approximately 10 minutes. Karen took Bus # 14 during peak hours on 36 different occasions. Her mean waiting time was 8.7 minutes. Assume that the population standard deviation o of 2.9 minutes is known. At the 0.01 significance level, test if the mean of all the peak hours waiting time for Bus # 14 is significantly different from 10 minutes.
- The heights are measured for a simple random sample of 9 supermodels. They have a mean of x= 70.0 inches and a standard deviation ofs= 1.5 inches. A data set of 40randomly selected women who are not supermodels has a mean ofx= 63.2 inches and a standard deviation ofs= 1.27 inches. Use a 0.01 significance level to test the claim that the mean height of supermodels is greater than the mean height of women who are not supermodels.The Ankle Brachial Index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. A healthy ABI is 0.9 or greater. In a study, researchers obtained the ABI of 187 women with arterial disease. The results were a mean ABI of 0.86 with a standard deviation of 0.16. At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, women with arterial disease have an unhealthy ABI? Set up the hypotheses for the one-mean t-test. Hou Ha: HA 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 4848 participants in the treatment group lowered their cholesterol levels by a mean of 21.521.5 points with a standard deviation of 2.52.5 points. The 4040 participants in the control group lowered their cholesterol levels by a mean of 20.920.9 points with a standard deviation of 4.14.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.100.10 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. step 3 of 3 : Conclusion- fail to reject/reject...insufficient/sufficient
- In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.2 and a standard deviation of 17.2. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? Ο Α. Hρ: μ=0 mg/dL O B. Ho: µ=0 mg/dL H;: u#0 mg/dL H,: µ>0 mg/dL O C . H: μ=0 mg/dL O D. Ho: µ>0 mg/dL H;: µ<0 mg/dL H,:µ<0 mg/dL Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value, (Round to three decimal…The General Electric Company clams that the life of LED light bulbs are has a mean life of 500 hours with the standard deviation of 50 hours. Test the claim of the manufacturer if a random sample of 80 LED light tested and found out to be 450 hours only. Use 5% level of significance.In a test of the effectiveness of garlic for lowering cholesterol, 4949 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.30.3 and a standard deviation of 2.142.14 Use a 0.100.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 00. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Upper H 0H0: muμequals=00 mg/dL Upper H 1H1: muμless than<00 mg/dL B. Upper H 0H0: muμequals=00 mg/dL Upper H 1H1: muμgreater than>00 mg/dL C. Upper H 0H0: muμequals=00 mg/dL Upper…
