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Eye Contact In a study of facial behavior, people in a control group are timed for eye contact in a 5-minute period. Their times are
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- Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4628 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2173. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 H_: 2 Ho H₂01 H: #02 2 2 02 2 022 202 Ho: 1 H₂ 2 1 02 Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. O Reject Ho. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers. Do not…arrow_forwardMileage data of new model cars follow a normal distribution with an average mileage of 26 mpg and a standard deviation of 7. Consider a sample of size 15 cars. What is the probability that the mean mileage of these selected cars is at most 27.5 mpg?arrow_forwardMethyl t-butyl ether (MTBE) is an organic water contaminant that often results from gasoline spills. The level of MTBE (in parts per billion) was measured for a sample of 12 well sites located near the ship channel during a recent spilI. The data are: 150, 376, 38, 12, 11, 134, 12, 251, 63, 8, 13, 107. (a) Find the sample mean and the sample standard deviation. (b) Construct and interpret a 91.85% and 98.15% confidence interval for the population mean by using the following equations: B, = x – t - - and By = x +t N sample N sample (c) Which confidence interval is wider and why?arrow_forward
- The patient recovery time from a particular surgical procedure was normally distributed with a mean of 5.6 days and a standard deviation of 2.1 days. What is the probability of spending more than three days in recovery? The 90th percentile for recovery time is?arrow_forwardIs there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4638 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2121. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 O Ho: o 2702 H: 01 Ho: 01 02 キ02 2ァ022 して 2. キ02 2502 2 %3D Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value %3D State your conclusion. O Reject Ho: We can conclude that there is a difference in the variability of golf scores for male and female professional golfers. O…arrow_forwardBone 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 womenarrow_forward
- what is the characteristis of the normal curve and what human trait or behavior is distributed normally?arrow_forwardA study was performed concerning risk factors for carotid-artery stenosis (arterial narrowing) among 264 men. The data reported is: Sample size Mean Standard deviationGroup 1: No stenosis 156 5.3 1.4Group 2: Stenosis 108 5.1 0.8 The test stastistic F is assumed to be 3.06 to check the equal variance. Question: Suppose researchers want to test if there is a significant difference in mean blood-glucose level between men with and without stenosis considering the critical F test is 1.35 with a equal variance (pooled), what is the test statistic for this test ?arrow_forwardIn 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…arrow_forward
- Low-fat or low-carb? Are low-fat diets more effective for weight loss? A sample of 46 subjects went on a low carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.6 kilograms with a sample standard deviation of 6.28 kilograms. A second sample of 49 subjects went on a low-fat diet. Their sample mean weight loss was 2.3 kilograms with a standard deviation of 4.66 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let μ1 denote the mean weight lost on the low-carb diet and μ2 denote the mean weight lost on the low-fat diet. Use the =α0.01level and the P-value method. Compute the test statistic. Round the answer to three decimal places. t =arrow_forwardA researcher decides to measure anxiety in group of bullies and a group of bystanders using a 23-item, 3 point anxiety scale. Assume scores on the anxiety scales are normally distributed and the variance among the group of bullies and bystanders are the same. A group of 30 bullies scores an average of 21.5 with a sample standard deviation of 10 on the anxiety scale. A group of 27 bystanders scored an average of 25.8 with a sample standard deviation of 8 on the anxiety scale. You do not have any presupposed assumptions whether bullies or bystanders will be more anxious so you formulate the null and alternative hypothesis based on that.arrow_forward
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