Using the p-value method, we reject the null hypothesis if the p-value is than or equal to the level of significance. 4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thi population produced a sample mean of 73.2. The population is normally distributed with o=6 a) Calculate the test statistic (round to two decimal places) b) Use z-table or technology to get the p-value (round to four decimal places) 5) Consider H u-56 versus Hy ut56 A random sample of 25 observations taken from this

Using the p-value method, we reject the null hypothesis if the p-value is than or equal to the level of significance. 4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thi population produced a sample mean of 73.2. The population is normally distributed with o=6 a) Calculate the test statistic (round to two decimal places) b) Use z-table or technology to get the p-value (round to four decimal places) 5) Consider H u-56 versus Hy ut56 A random sample of 25 observations taken from this

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Description: Using the p-value method, we reject the null hypothesis if the p-value isthan or equal to the level of significance.4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thispopulation produced a sample mean of 73.2. T

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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A sample of 20 elements selected from a population produced a mean of 65 and a standard deviation of 9. Another sample of 25 elements selected from another population produced a mean of 58 and a standard deviation of 11. Assume that the two populations are normally distributed and the standard deviations of the two populations are equal. The null hypothesis is that the means of the two populations are equal and the alternative hypothesis is that the means of the two populations are not equal. The significance level is 5%. The critical values of t are: a. –2.014 and 2.014 b. –2.017 and 2.017 c. –1.681 and 1.681 d. –1.679 and 1.679
A random sample of 54 students from Cabrillo College are selected from a normal population with mu equal to 20 and sigma equal to 5. After a treatment is administered to the students, the researcher calculated the sample mean to be 21. Is there enough evidence to conclude that the treatment had an effect at the 5% significance level? If the z-statistic equals 1.46 and the critical values are -1.96 and 1.96, what conclusion can we make? (a)reject the null hypothesis and accept the alternative hypothesis (b)cannot determine (c)fail to reject the null hypothesis (d)fail to reject the alternative hypothesis
1)Remove the four potential outliers of 0, 0, 8, and 20, and then obtain a new histogram without the outliers. Does the data appear to be normally distributed now? 2)Assuming that the four potential outliers of 0, 0, 8, and 20 are not recording errors, repeat the hypothesis test from part (c) (again setting up the hypothesis test and using either the critical value or p-value approach), and compare your results with that obtained in (c). Did you make a different conclusion? 3)Imagine you know have to make a recommendation/conclusion to the company that hired you: Assuming that the four potential outlies are not recording errors, and looking at the two results above, would you recommend using the first test with the outliers or the second test with the outliers removed? There is no right or wrong answer here, I am interested in what you think and your reasoning.
Follow the steps in Test of Hypothesis (pic) including your conclusion and round off the critical value and test statistic 4 decimal places Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measures in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in x1 = 290 and s1 = 12 while another random sample of 16 gears from the second supplier results in x2 = 321 and s2 = 22. Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength. Use a =0.04, and assume that both populations are normally distributed with equal variances.
An investigator is interested if there is a difference in the time (in seconds) to sprint a given distance among non-smokers, past smokers and current smokers. He obtained the following data: 1. State your hypotheses. 2. What is the appropriate statistical test to determine if there is a difference in the sprint times among different smoking groups? 3. The investigator uses α = 0.01 as the level of significance (df = 2, 350). Give the critical value and region of rejection. 4. Should the null hypothesis be rejected? 5. State your conclusion. 6. Given your conclusion, what is the next procedure that should be done? 7. If the investigator classified the smoking groups according to their gender and obtained the mean sprint times per gender for each group, what type of statistical test should be used now?
A major credit card company is interested in the proportion of individuals who use a competitor’s credit card. Their null hypothesis is H0: p=0.65H0: p=0.65, and based on a sample they find a sample proportion of 0.70 and a pp-value of 0.053. Is there convincing statistical evidence at the 0.05 level of significance that the true proportion of individuals who use the competitor’s card is actually greater than 0.65 ?
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A major credit card company is interested in the proportion of individuals who use a competitor’s credit card. Their null hypothesis is H0: p=0.65H0: p=0.65, and based on a sample they find a sample proportion of 0.70 and a pp-value of 0.053. Is there convincing statistical evidence at the 0.05 level of significance that the true proportion of individuals who use the competitor’s card is actually greater than 0.65 ? Yes, because the sample proportion 0.70 is greater than the hypothesized proportion 0.65. A Yes, because the pp-value 0.053 is greater than the significance level 0.05. B No, because the sample proportion 0.70 is greater than the hypothesized proportion 0.65. C No, since the sample proportion 0.70 is exactly 0.05 away from the hypothesized proportion 0.65. D No, because the pp-value 0.053 is greater than the significance level 0.05.
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To determine whether training in a series of workshops on creative thinking increases IQ scores, a total of 70 students are randomly divided into treatment and control groups of 35 each. After two months of training, the sample mean IQ for the treatment group equals 110, and the sample mean IQ for the control group equals 108. The estimated standard error equals 1.80. Using t, test the null hypothesis at the .01 level of significance. If appropriate (because the null hypothesis has been rejected), estimate the standardized effect size, construct a 99 percent confidence interval for the true population mean difference, and interpret these estimates.