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- st • Question 14 You wish to test the following claim (Ha) at a significance level of a = 0.002. H.:p = 51.2 H.:p > 51.2 You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: data 69 58.6 62.8 sUcfocthis sampls/Repout.answeraccurato three ecCan you help me with the p-value I answered 0.034 which is incorrect???
- A researcher believes that low BMI levels are indicators of eating disorders. He collects BMI of a sample of young adults identified as at risk for eating disorfers to see if their mean BMI is significantly below 18.5 What test would you use and why? (z test, independent t test, single sample t test, paired samples)Microsoft Word -... # The average number of accidents at controlled intersections per year is 5.4. Is this average more for intersections with cameras installed? The 47 randomly observed intersections with cameras installed had an average of 5.9 accidents per year and the standard deviation was 1.08. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: V = H₁: > 5.4 c. The test statistic t = 3.174 (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? a f. Based on this, we should g. Thus, the final conclusion is that ... 80 F3 V 5.4 Select an answer the null hypothesis. O The data suggest that the populaton mean is significantly more than 5.4 at a = 0.01, so there is statistically significant evidence to conclude that the population mean number of accidents per year at intersections with…An overly involved "neighborhood watch" group has been investigating the length of lawns. In years past, lawn lengths are normally distributed and lawns had been mowed to a mean length of 3.5 inches. A recent random sample of 16 lawn lengths is given below. Conduct an appropriate test, at a 5% significance level, to determine if the overall mean lawn length is now higher than 3.5 inches. Round all answers to 4 decimal places. Lawn Length (in.) 3.14 3.64 3.49 3.04 4.09 4.54 3.86 4.27 5.01 3.9 4.77 3.54 3.97 2.51 4.08 3.39 Checksum: 61.24 a. Consider the parameter of interest and choose the correct alternate hypothesis. μ≠3.5μ≠3.5 μ<3.5μ<3.5 μ=3.5μ=3.5 μ>3.5μ>3.5 b. Are the necessary conditions present to carry out this inference procedure? Explain in context. Our sample was not larger than 30 in size, but it is stated that the sample was randomly gathered from a normally distributed population. No. The sample is not large…